Given enough time, everyone is capable of computing if say ; all operations reduce to addition/multiplication of rational numbers. But have you ever asked yourself how you actually compute say or or maybe ? It turns out that we have no way to compute exactly (and similarly and ), even with the world's fastest super computer with all the available memory. We can, however, approximate such quantities and even determine a bound on the error of our approximation. How, you ask? Well, one of the main goals of this part of the course is to show that many familiar functions can be represented as an infinite sum of polynomials. For example, we will show that for all it is true that One of the uses of representing a function with an infinite sum of polynomials is that we can truncate the infinite sum to obtain an approximation using a finite sum of polynomials. For example, How else do you think your calculator computes ? Consider the expansion for . Since , then by plugging in in the representation of we obtain that and therefore multiplying both sides by we obtain Or consider the expansion for . Plugging in into the representation of we obtain Before we can understand how to represent a function as a sum of infinite polynomials, we need to understand what it means for an infinite sum of real numbers to ``sum-up'' to a finite number. Infinite sums are called infinite series or just series, and to understand series we need to understand convergence of sequences.

Sequences

A sequence is a list of numbers in a given order: Here are some sequences: The th term of a sequence is denoted by . For instance, in the last sequence above, , , , and so on. The integer is called the index of the sequence and usually the index starts at and sometimes at . A sequence may be defined using a formula for the general th term . For example, the sequence of even numbers has general th term The sequence has general th term Notice that for this last sequence, we could write the general term in the form which is aesthetically more pleasing to the eye.
Write out the first few terms of the given sequence
Some general notation you might see for denoting a sequence is , , , etc. From now on we will use .

Convergence of Sequences

Before we discuss convergence of sequences, we review some basics regarding absolute value and inequalities. First of all, the inequality means that , or lies in the interval . Given numbers and , the number is the distance between and . Therefore, if then the distance between and is less than : and adding to all terms Hence, lies in the interval . Consider the sequence with th term . The first few terms are Here is a picture:
figures/sequence-on-real-line.svg
We will show that for sufficiently large, the numbers get arbitrarily close to . In other words, for any given small quantity , the terms of the sequence from some point and onward will lie in the interval In other words, for sufficiently large it holds that For example, if say then for which is it true that ? Consider first Now when Therefore, if then or . In fact, If instead then need for to lie in the smaller interval . However, no matter how small is, there exists some natural number such that if then lies in the interval .
A sequence is said to converge if there exists a number such that for any given there exists a natural number such that if then . In this case, we then write that and call the limit of .
Show by definition that the sequence converges to . Find the smallest number such that if then .
Let be arbitrary. Now Now when Hence, need . The minimal is thus .
A sequence is said to be bounded if all the terms of the sequence are contained in some interval, in other words, there exists such that for all . If is bounded then does necessarily converge? Not necessarily, for example consider the sequences If is convergent then is necessarily bounded? Yes! For any , there exists such that for all . The other terms may not be in the interval but there is only finitely many of them. As an example, consider Then and this gets arbitrarily large and thus is not bounded; therefore does not converge. To compute a limit we can compute it as if we were computing a functional limit of the form and we can even use l'H\^{o}pital's rule. Given sequences and , we can obtain a new sequence given by . In other words, We can also define the difference, product, division, and scalar multiplication of sequences:
Suppose that and exist. Then:
  5. If and then
Compute the following limits.
  1. (Do it by factoring )
  2. (Bring limit inside)
  3. (Bring limit inside)
  4. (H.R.)
We must be careful how we apply the Limit Laws. For example, consider . Then we might be tempted to write However, does not exist and thus we cannot apply the Limit Laws. In this case, we must use the following.
Suppose that for all . If and then .
Compute the following limits.
  1. (Squeeze)
  2. (Squeeze)
Find
Get an indeterminate form of the type . Consider then Hence, .
Find
Write as Because then . Hence, . In fact, for any number it holds that If then does not exist.
A sequence is increasing if and is called decreasing if A sequence that is either increasing or decreasing is called monotone. If is convergent is it necessarily increasing or decreasing? Not necessarily, for example consider the sequence or It is fairly clear that but is not monotone. We now know that a sequence that is bounded is not necessarily convergent. But what about a sequence that is monotone and bounded?
If is bounded and monotone then is convergent.
Consider the sequence whose general th term is Notice that It is now clear that for any . Therefore, is increasing. It can be shown that is bounded. Therefore, by the MCT, exists.

Infinite Series

A sequence gives rise to a new sequence called the sequence of partial sums of defined as follows: Using summation convention, the th term of is therefore Notice that
Let for . Find the first ten terms of the sequence of partial sums associated to .
We compute In fact, with a bit more patience one can compute that It seems like .
We will be interested in computing limits of sequences of partial sums. We define an infinite series as the limit of the sequence of partial sums of a given sequence, and we use the following notation for the limit: Hence, our previous example suggests that or using expanded notation In general, if exists then we say that the infinite series converges, otherwise we say that it diverges.
Determine whether the infinite series generated by the sequence converges.
Here and thus the sequence of partial sums is Here are the first few terms of : It seems as though is divergent because it is unbounded. To see this: and thus For instance, and agrees with our previous computation. It is clear that does not exist. Hence, the infinite series diverges.
Let be a constant and let for . Investigate the convergence of the infinite series generated by .
The infinite series is The sequence of partial sums has general th term We can get a closed-form formula for by using and thus we obtain that We need to consider the limit , which amounts to determining : Therefore, the series converges when and it converges to This series is the most important series in this course; it is called the GEOMETRIC SERIES. Notice that we have
Determine whether the given series converges, and if yes, determine the value of the convergent series if possible.
  4. (DNE)
Use a geometric series to show that
Below is a test for divergence of a series.
Consider the series . If then the series diverges. In other words, if converges then it must hold that .
Using the divergence test, we can easily deduce that following series diverge:
In examples (i)-(ii), the limits do not exist. In case (ii), , and in case (iv) .
Consider the series In this case, and . However, we cannot conclude that the given series converges. In fact, it can be shown that the sequence of partials sums associated to the sequence is unbounded. Therefore, diverges and consequently the given series diverges.
We now state some basic arithmetic properties of infinite series.
Suppose that and are convergent. Then
  1. converges and
  2. converges and
  3. For any constant , converges and
Important: We note that if converges and diverges then diverges.
Determine if the given series converge and if possible find the limit.
  1. :
  2. : Now and thus the series diverges by the Divergence test.
  3. : We have that and thus the series diverges by the Divergence test.
  4. : Notice that , , , and . Hence, . Therefore
  5. : First notice that and thus the series may converge. Now Since then the geometric series converges to
  6. : This resembles a geometric series except the initial value of the index is . However, we can write An alternative method is to expand the series:
A particular series has sequence of partial sums whose th general term is . (5 minute group work, 3 students give answer on board)
  1. Find and .
  2. Does the series converge? Explain. If yes, what does it converge to?
  3. Does the sequence converge? Explain. If yes, what does it converge to?
One last note:

Integral Test

So far, the only series we have been able to show converges is the Geometric series provided , and any of its variants. This was possible because we were able to show that and then . In the next few sections, we will develop general tests to deduce the convergence of other series. Our first general test is the Integral test and relies on the Monotone Convergence theorem. Recall that given a series if the sequence of partial sums is increasing and bounded above then converges by the MCT and thus converges. We now determine under what circumstances is increasing. Recall that by definition and thus If the sequence has all non-negative terms, that is, for all , then and thus is increasing. Thus, in this case, converges if it is bounded above. Consider the series . Since then is increasing. If is bounded above then converges. We can view the series as an infinite Riemann sum:
figures/integral-test.svg
Compare the area under the function from to : But and therefore The improper integral involved converges Hence, Therefore, is bounded above! By the Monotone Convergence theorem, the series converges, but we cannot deduce the sum. We write down the general statement of the Integral Test.
Let be a sequence with and suppose , where is non-negative, continuous, and decreasing. Then is convergent if and only if is convergent.
Let and consider the series . Hence where . The function is continuous, non-negative, and decreasing on the interval . Now and then Therefore, the general -series converges if and diverges otherwise. For example, , converge but diverges.
Determine if the given series converge or diverge.
  1. : In this case, where , and is continuous, decreasing, and non-negative on . Now Therefore, by the integral test, the given series converges because the improper integral converges.
  2. : First notice that and therefore the series may converge. In this case, where , and is continuous and non-negative on the interval . Now when and thus is decreasing on . Now Therefore, since the improper integral converges, then by the Integral test the given series converges.
  3. : First notice that and thus the sequence may converge. Now where , and is continuous, non-negative, and decreasing on . Now Hence, since the improper integral diverges then the given series diverges.

Comparison Tests

Thus far, we know that
  1. if ,
  2. diverges, and
  3. converges only if .
We now introduce Comparison tests which allow us to deduce either convergence/divergence of a given series when we know the convergence/divergence of another series and when .
Suppose that and are sequences of non-negative terms: and . Suppose further that for all . The following hold:
  1. If converges then converges.
  2. If diverges then diverges.
Notice that this theorem does not say what happens when and when converges or diverges. In what follows, the following fact will be used frequently. If and then : Also, if then also
First , hence Divergence test is inconclusive. Also, for all . When is very large, and thus Hence, when is large the terms of the series are like the terms of the Harmonic series. We have reason to believe then that the given series diverges. Now because then and thus The series diverges and thus by the DCT the given series diverges.
We have that and . When is large Thus, the terms of the series behave like the terms of the series ; we have reason to believe that the given series is convergent. Now, Hence, . The series converges and thus by the DCT, the given series converges also.
Here . The following is true: and in fact Below is a graph:
figures/log-sqrt.svg
Therefore, The -series converges and thus by the DCT test the given series converges.
Here for all , and when is large Thus, there is reason to believe that the series diverges. Now Thus, . Since the series diverges then by the DCT test, the given series diverges.
We see that applying the Direct Comparison test requires some ingenuity. The following theorem is derived from the Direct Comparison test and some cases is easier to apply than the DCT.
Suppose that and for all .
  1. If exists and then and have the same convergence/divergence property, i.e., either both converge or both diverge.
  2. If and if converges then converges also.
  3. If and if diverges then also diverges.
Usually, we know the convergence/divergence properties of and we are interested in knowing the convergence/divergence properties of . Thus we compute the limit and apply the LCT. However, the difficult part with the LCT is to first determine the correct series to compare with and this is done using the preliminary analysis. Let's redo the previous examples with the LCT.
As before, we is large Now, diverges and Hence, by the LCT, the given series diverges.
As before, when is large The series converges and Thus, by the LCT, the given series converges. Suppose we compare with , which converges. Now Then since converges then the given series also converges.
Suppose we blindly compare with , which diverges: When the limit is zero, we can only conclude that converges if converges; but does not converge and thus the test is inconclusive using . If we try with , which converges, we get In this case, part (iii) of the LCT is not applicable because converges. In this case, the DCT is actually easier because we know the bound and then and converges. Note that if we try the LCT with which converges then Hence, part (ii) of the LCT is applicable and we conclude that the given series is convergent.
Here is negative for and thus cannot apply the comparison tests directly. However, notice that The second series is geometric with and thus converges. For the first series we have and thus by the D.C. the series converges. Hence, the given series converges.
Notice that The -series converges and thus by the D.C. the given series converges.
When is large and so we have reason to believe the series diverges. Now Since diverges then the given series diverges. Alternatively, could use limit comparison test with .

Absolute Convergence and Ratio Test

Consider the series This is called an alternating series. The series is clearly not geometric, we cannot apply integral test or any comparison test because the series contains negative terms. However, we know that converges. Does this imply that the original series converges? Well and thus If converges then converges by the Direct Comparison test. Therefore, by the series laws converges also.
Let be any given series. If converges then converges. In this case, we say that converges absolutely.
Because has non-negative terms, we can apply the integral test or any of the comparison tests to , when they are applicable.
We can then conclude that the series converges absolutely because is a convergent -series.
Consider the series . Here we have Applying L.H. rule, we get The exponential grows much faster than , so suspect series converges. Let then by applying L.H. Rule. Hence, since converges then converges. And then by Absolute convergence test, the given series converges absolutely. Note that would have also worked.
Have negative terms so consider Then with we obtain Hence, by LCT, the series converges because converges. Then by ACT, the original series converges absolutely.
Have negative terms so consider Suspect converges but if we try the LCT with say we get and this limit DNE. Thus, the LCT is inconclusive. However, notice that and converges and thus converges by the DCT. Thus, the given series converges absolutely.
We now introduce a very powerful convergence test.
Let be any given series with and let
  1. If then converges absolutely.
  2. If or then diverges.
  3. If or the limit does not exist then the test is inconclusive.
Consider again . We compute and then and thus by the Ratio Test, the series converges absolutely.
Compute and thus Hence, the series converges by the RT.
Compute and thus and thus by the RT the given series diverges.
We compute and therefore and thus by the RT the given series converges.
We compute and thus and thus by the RT the given series diverges.
The Ratio Test doesn't always work or can be difficult to apply. Consider for instance In the first case, we get and the limit DNE. Here, it is easier to apply absolute convergence combined with the DCT. In the second case, the limit computation is just tedious but does give . Again, could just apply ACT combined with DCT.

Alternating Series

Consider the series The ratio test is inconclusive because and the series of the absolute values is divergent and so the ACT is inconclusive. The given series is called the alternating Harmonic series. It turns out that this series converges!
The alternating series converges if the following hold:
  1. for all
  2. is decreasing or eventually decreasing
For the alternating series, , we have and all three conditions are satisfied and thus the alternating harmonic series converges. However, because diverges we also say that converges conditionally.
Trying the ratio test gives and the series of absolute values diverges, so the ACT is inconclusive. In this case, and . Now we get that . To show this, let . Then Now if and only if and . Hence, is decreasing starting with . All three conditions are met and thus by the Alternating series test the given series converges conditionally.
Again, the RT test and ACT are inconclusive for the same reasons as the previous example. In this case, and clearly . Now let . Then iff . So, decreasing for . Thus, the series converges conditionally.

Power Series (Part I)

Recall that provided . Think of as a variable that varies in the interval , and call it instead: Hence, in the interval , the function can be represented as an infinite series. So what? If we can represent in this way maybe we can do the same for other more complicated functions. Notice the following: Does this mean then that Not clear that this is valid. What is ? At we have and thus . Hence Therefore, if then Notice then that What is the main idea?
A power series is a series of the form where the number is called the center of the power series and the terms of the sequence are called the coefficients of the power series. Using summation notation: When the center is then
Expand the given power series by writing out the first few terms of the series.
Use summation notation to write the given power series.
Given a power series, the first question we would like to answer is for which values of does the series converge.
Find all the values of for which the given power series converges.
Here the center is . Suppose we set : Apply the Ratio test: Thus, when , the resulting series converges. If instead we set : and in this case the Ratio test gives and thus the resulting series diverges when . Apply the Ratio test for general : By the Ratio test, the series converges when or when , or when . Note that the R.T. is inconclusive when and when the series diverges. The case occurs when which happens at the boundary points and . We need to check these manually. At we obtain which diverges. On the other hand, at the boundary point we obtain which converges. Hence, the interval of convergence is . We can therefore define the function whose domain is .
Find the values of where the given power series converges.
Applying the Ratio test yields Hence, by the Ratio test, the series converges if or when . When , the Ratio test is inconclusive, which occurs when or . At we obtain which diverges and at we obtain which also diverges. Hence, the interval where the series converges is .
For any given power series , one of the following will hold:
  1. The series converges for satisfying where is a constant, that is, when . At the boundary points , the series may or may not converge. Outside the interval, that is, when , the series diverges.
  2. The series converges for any , that is, the interval of convergence is .
  3. The series converges only for .
The number is called the radius of convergence and the interval (with possibly one or both of the boundary points included) is called the interval of convergence. In case (ii), we could say that and in case (iii) that .
For each power series, determine the the interval and radius of convergence.
  2. Only for ,
  3. For ,
  4. For , . Graph the resulting function on the domain .
Find the interval of convergence of and graph the resulting function on its domain.

Power Series (Part II)

We know that the function can be represented as a power series in the interval because In this case, and the center is . In this section, we are going to use this power series representation of to construct power series representations of other functions. Representing a function as a power series has many applications and we will see some applications soon. Suppose that is a function defined for satisfying , that is, in the interval . To give an example, suppose that and thus . Let and consider the new function . What is the domain of ? The function is well-defined if Hence, is defined when or on the interval .
Determine a power series representation for the given function and determine the interval where the representation is valid.
We know that and this is valid for or for . Notice that and this is valid for or for .
Find the interval of convergence of the given power series and find a closed-form expression for the function given by the power series.
Notice that This is a geometric series with and we know a geometric series converges when : Hence, interval of convergence is and the closed-form expression is
Find a power series representation for the given function and determine the interval where the representation is valid. What is and what is ?
Notice that and the representation is valid when : Hence, the interval of convergence is . The power series we found is of the form In our representation, the polynomial appears when , hence and since all powers of appearing in the series are odd then .
How do we differentiate/integrate functions that are represented as power series?
Suppose that with interval of convergence . Then the function is continuous and differentiable on the interval . Moreover, to find and as a power series we simply differentiate/integrate the power series of : And
When we integrate, the new power series may now converge at the boundary points, and conversely, when we differentiate we may lose convergence at the boundary points. This must be checked on a case-by-case basis.
Find a power series representation for and determine where the representation is valid. Use your power series to show that
We know that Now and this representation is valid when or . Therefore, To find we evaluate both sides at a convenient point, say and this gives . Hence, and this is valid for . However, notice that at we get which by the Alternating series test converges. Hence, and since we get
Find a power series representation for and the interval where the representation is valid.
If we can find a power series representation for then we will need to multiply the series by to obtain a power series for . Now, so we will find a power series for and then integrate to get a power series for . Now valid for or for . Then and valid for . Now, evaluating both sides at gives . Hence
Find a power series representation for and where the representation is valid.
We know that valid in the interval . Now, Therefore, and thus and the representation is valid in the interval .
Find a power series representation for the given function and where the representation is valid If what is and ?
Compute directly Since then and the interval is . The power does not appear and thus . In our representation, the term corresponds to and thus .
Show that and determine where the representation is valid. Use your representation to show that

Taylor Series

We saw how we could use the geometric series to find power series representations for functions such as , , or . We now study a systematic alternative method to find power series representations for any reasonably behaved function. Suppose that is a given function in closed-form and we wish to represent it as a power series: Because a power series is determined by its coefficients, to find a power series for we need to find . First of all, notice that Now consider and thus Similarly, and thus Similarly, and thus This pattern continues: Thus, if is represented as a power series centered at then it must be This power series is called the Taylor series of centered at . If , then the series is called the Maclaurin series, so in this case:
Recall that we found a power series representation for : This was obtained by integrating the power series for . Let's compute the Maclaurin series of to compare: Therefore, the Maclaurin series is which is exactly what we computed when using the geometric series.
Find the Maclaurin series of and determine its interval of convergence. What is , the 8th order Taylor polynomial centered at ?
We compute and the pattern repeats. We get Therefore, Performing the Ratio test on the Maclaurin series we get and thus the interval of convergence is . Therefore,
Compute the Maclaurin series of and find its interval of convergence.
We could proceed as was done for or we could use the fact that . Differentiating the power series of we obtain with interval of convergence .
Find the Maclaurin series of and determine its interval of convergence. What is , the 4th order Taylor polynomial centered at ?
We need to compute . But since for all then . Thus the Maclaurin series is And thus To find the interval of convergence compute and thus the interval of convergence is . Therefore, This shows, for instance, that
Find a power series representation for centered at .
Compute The pattern is . Therefore, the series centered at is
Suppose that and valid on the interval . To approximate the infinite sum, we could use finite sum This is called the th order Taylor polynomial of centered at . When is close to the center , then The higher the , the more accurate the approximation is.
Consider the function
  1. Find the Maclaurin series of and find its interval of convergence.
  2. Find .
  3. Find .
  4. Use to estimate .
Compute Because the power series of converges on then the Taylor series of converges on . To find recall that a function's power series is its Taylor series, so if then Then and thus Now from our power series representation, the term corresponds to and thus Then Now to find we expand and keep only up to the term if present: and therefore Then
Find a power series representation, centered at , for What is ?
We have Then
Find the Taylor series of at . Then find .
After successive differentiation one finds that and thus . Thus the Taylor series is and then
Find the Maclaurin series of
Not every function is equal to its Taylor series, that is, it isn't always true that For example, the function has derivatives at all equal to zero, . Then the Taylor series is This power series then converges on . However, it is clear that is not identically equal to zero:
figures/bump-function.svg