3. Sequences and Series

Let \(\eps \gt 0\) be arbitrary. Now \[ |a_n - \tfrac{1}{2}| = \left|\frac{2n-2-2n-1}{2n+1} \right| = \left|\frac{-3}{2(2n+1)}\right| = \frac{3/2}{2n+1} \] Now \[ \frac{3/2}{2n+1} \lt \eps \] when \[ \frac{3}{4\eps}-\frac{1}{2} \lt n \] Hence, need \(M \gt \frac{3}{4\eps}-\frac{1}{2}=7499.5\). The minimal \(M\) is thus \(M=7500\).

Get an indeterminate form of the type \(1^\infty\). Consider then \begin{align*} \limi \ln \left(\frac{n+1}{n-1}\right)^n &= \limi n \ln \left(\frac{n+1}{n-1}\right)\\ &= \limi \frac{\ln \left(\frac{n+1}{n-1}\right)}{1/n} \\[2ex] &= \limi \frac{\frac{1}{n+1}-\frac{1}{n-1}}{-1/n^2}\\[2ex] &= \limi \frac{2n^2}{n^2-1} = 2 \end{align*} Hence, \(\ds\limi \left(\frac{n+1}{n-1}\right)^n = e^2\).

Write as \begin{align*} \limi \left(\frac{3}{4}\right)^n &= \limi e^{\ln (3/4)^n}\\ &= \limi e^{ \ln(3/4) n} \end{align*} Because \(\ln(3/4) \lt 0\) then \(\ds\limi e^{\ln(3/4)n} = 0\). Hence, \(\ds\limi (3/4)^n = 0\). In fact, for any number \( -1 \lt p \lt 1\) it holds that \[ \limi p ^n = 0. \] If \(|p| \gt 1\) then \(\ds\limi p^n\) does not exist.

We compute \begin{align*} s_1 &= a_1 = 1\\ s_2 &= a_1 + a_2 = 1 + \tfrac{1}{2} = \tfrac{3}{2}=1.5\\ s_3 &= a_1 + a_2 + a_3 = \tfrac{3}{2} + \tfrac{1}{4} = \tfrac{7}{4}=1.75\\ s_4 &= \tfrac{7}{4} + \tfrac{1}{8} = \tfrac{15}{8}=1.875\\ s_5 &= s_4 + \tfrac{1}{16} = 1.9375\\ s_6 &= s_5 + \tfrac{1}{32} = 1.96875\\ s_7 &= s_6 + \tfrac{1}{64} = 1.984375\\ s_8 &= s_7 + \tfrac{1}{128} = 1.9921875\\ s_9 &= s_8 + \tfrac{1}{256} = 1.99609375\\ s_{10} &= s_9 + \tfrac{1}{512} = 1.998046875 \end{align*} In fact, with a bit more patience one can compute that \begin{align*} s_{20} &= 1.9999980926513672\\ s_{50} &= 1.9999999999999982 \end{align*} It seems like \(\ds\lim_{k\rightarrow\infty} s_k = 2\).

Here \(a_n = n\) and thus the sequence of partial sums is \[ s_k = a_1 + a_2 + \cdots + a_k = 1 + 2 + 3 + \cdots + k \] Here are the first few terms of \(s=(s_1,s_2,s_3,s_4,\ldots)\): \[ s_1 = 1,\; s_2 = 3,\; s_3 = 6,\; s_4 = 10,\; s_5 = 15 \] It seems as though \(s\) is divergent because it is unbounded. To see this: \begin{align*} s_k &= 1 + 2 + 3 + \cdots + k \\ s_k &= k + (k-1) + (k-2) + \cdots + 1 \end{align*} and thus \[ 2 s_k = k(k+1)  \Longrightarrow   s_k = \frac{k(k+1)}{2} \] For instance, \(s_5 = 5\cdot 6 / 2 = 15\) and agrees with our previous computation. It is clear that \[ \limik s_k = \limik \frac{k(k+1)}{2} \] does not exist. Hence, the infinite series \(\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty n\) diverges.

The infinite series is \[ \sum_{n=0}^\infty r^{n} = 1 + r + r^2 + r^3 + \cdots \] The sequence of partial sums \((s_n)\) has general \(k\)th term \[ s_k = \sum_{n=0}^k r^{n} = 1 + r + r^2 + \cdots + r^{k}. \] We can get a closed-form formula for \(s_k\) by using \[ (1-r)(1+r+r^2+\cdots+r^{k}) = 1-r^{k+1} \] and thus we obtain that \[ s_k = 1+r+r^2+\cdots+r^{k} = \frac{1-r^{k+1}}{1-r} \] We need to consider the limit \(\ds\lim_{k\rightarrow\infty} s_k\), which amounts to determining \(\ds\lim_{k\rightarrow\infty} r^k\): \[ \lim_{k\rightarrow\infty} s_k = \lim_{k\rightarrow\infty} \frac{1-r^{k+1}}{1-r} = \begin{cases} \frac{1}{1-r}, & |r| \lt 1\\[2ex] \text{DNE}, & |r|\geq 1 \end{cases} \] Therefore, the series \(\sumz r^{n}\) converges when \(|r| \lt 1\) and it converges to \[ \sumz r^{n} = \frac{1}{1-r}. \] This series is the most important series in this course; it is called the GEOMETRIC SERIES. Notice that we have \[ \sumz r^n = \sumo r^{n-1} \]

First \(\ds\limi \frac{5}{5n-1} = 0\), hence Divergence test is inconclusive. Also, \(a_n \geq 0\) for all \(n\). When \(n\) is very large, \(5n-1\approx 5n\) and thus \[ \frac{5}{5n-1} \approx \frac{5}{5n} = \frac{1}{n}. \] Hence, when \(n\) 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 \(5n-1 \lt 5n\) then \(\frac{1}{5n} \lt \frac{1}{5n-1}\) and thus \[ \frac{5}{5n} \lt \frac{5}{5n-1}. \] The series \(\sumo \frac{5}{5n} = \sumo \frac{1}{n}\) diverges and thus by the DCT the given series diverges.

We have that \(a_n = \frac{n+3}{2n^4+5}\geq 0\) and \((a_n)\rightarrow 0\). When \(n\) is large \[ a_n = \frac{n+3}{2n^4+5} \approx \frac{n}{2n^4} = \frac{1}{2n^3}. \] Thus, the terms of the series behave like the terms of the series \(\sumo \frac{1}{2n^3}\); we have reason to believe that the given series is convergent. Now, \[ \frac{n+3}{2n^4+5} \lt \frac{n+3}{2n^4} \leq \frac{n+3n}{2n^4} = \frac{2}{n^3}. \] Hence, \(a_n \leq \frac{2}{n^3}\). The series \(\sum \frac{2}{n^3}\) converges and thus by the DCT, the given series converges also.

Here \(a_n = \frac{\ln(n)}{n^2}\). The following is true: \[ \ln(x) \leq \sqrt{x} = x^{0.5} \] and in fact \[ \ln(x) \leq x^{0.4}. \] Below is a graph:

Therefore, \[ \frac{\ln(n)}{n^2} \leq \frac{n^{0.4}}{n^2} = \frac{1}{n^{1.6}} \] The \(p\)-series \(\sumo \frac{1}{n^{1.6}}\) converges and thus by the DCT test the given series converges.

Here \(a_n = \frac{n+2}{n^2-n}\geq 0\) for all \(n\geq 2\), and when \(n\) is large \[ a_n = \frac{n+2}{n^2-n} \approx \frac{n}{n^2} = \frac{1}{n}. \] Thus, there is reason to believe that the series diverges. Now \[ a_n = \frac{n+2}{n^2-n} \geq \frac{n+2}{n^2} \gt \frac{n}{n^2} = \frac{1}{n}. \] Thus, \(\frac{1}{n} \leq a_n\). Since the series \(\sum_{n=2}^\infty \frac{1}{n}\) diverges then by the DCT test, the given series diverges.

As before, we \(n\) is large \[ a_n = \frac{5}{5n-1} \approx \frac{5}{5n} = \frac{1}{n}. \] Now, \(\sumo b_n = \sumo \frac{1}{n}\) diverges and \[ \limi \frac{a_n}{b_n} = \limi \frac{5n}{5n-1} = 1 \gt 0. \] Hence, by the LCT, the given series diverges.

As before, when \(n\) is large \[ a_n = \frac{n+3}{2n^4+5} \approx \frac{n}{2n^4} = \frac{1}{2n^3}. \] The series \(\sumo b_n = \sumo \frac{1}{n^3}\) converges and \[ \limi \frac{a_n}{b_n} = \limi \frac{n^4 + 3n^3}{2n^4 + 5} = \frac{1}{2} \gt 0. \] Thus, by the LCT, the given series converges. Suppose we compare with \(\sumo b_n = \sumo \frac{1}{n^2}\), which converges. Now \[ \limi \frac{a_n}{b^n} = \limi \frac{n^3+3n^2}{2n^4 + 5} = 0. \] Then since \(\sumo b_n\) converges then the given series also converges.

Suppose we blindly compare with \(\sumo b_n = \sumo \frac{1}{n}\), which diverges: \[ \limi \frac{a_n}{b_n} = \limi \frac{n \ln(n)}{n^2} = \limi \frac{\ln(n)}{n} = 0. \] When the limit is zero, we can only conclude that \(\sumo a_n\) converges if \(\sumo b_n\) converges; but \(\sumo b_n\) does not converge and thus the test is inconclusive using \(\sumo \frac{1}{n}\). If we try with \(\sumo \frac{1}{n^2}\), which converges, we get \[ \limi \frac{a_n}{b_n} = \limi \ln(n) = \infty. \] In this case, part (iii) of the LCT is not applicable because \(\sumo \frac{1}{n^2}\) converges. In this case, the DCT is actually easier because we know the bound \(\ln(n) \leq \sqrt{n}\) and then \[ \frac{\ln(n)}{n^2} \leq \frac{n^{0.5}}{n^2} = \frac{1}{n^{1.5}} \] and \(\sumo \frac{1}{n^{1.5}}\) converges. Note that if we try the LCT with \(\sumo b_n = \sumo \frac{1}{n^{1.5}}\) which converges then \begin{align*} \limi \frac{a_n}{b_n} = \limi \frac{n^{1.5}\ln(n)}{n^2} \limi \frac{\ln(n)}{n^{0.5}} &= \limi \frac{(1/n)}{\frac{1}{2}n^{-1/2}} \\ &=\limi \frac{2}{n^{1/2}} = 0 \end{align*} Hence, part (ii) of the LCT is applicable and we conclude that the given series is convergent.

Here \(a_n = \frac{1-n}{n2^n}\) is negative for \(n \gt 1\) and thus cannot apply the comparison tests directly. However, notice that \[ \sumo \frac{1-n}{n2^n} = \sumo \frac{1}{n2^n} - \sumo \frac{1}{2^n} \] The second series is geometric with \(|r|=1/2 \lt 1\) and thus converges. For the first series we have \[ \frac{1}{n2^n} \leq \frac{1}{2^n} \] and thus by the D.C. the series \(\sumo \frac{1}{n2^n}\) converges. Hence, the given series converges.

Notice that \[ \frac{n^2}{\sqrt{n^9+2}} \leq \frac{n^2}{\sqrt{n^9}} = \frac{1}{n^{2.5}} \] The \(p\)-series \(\sumo \frac{1}{n^{2.5}}\) converges and thus by the D.C. the given series converges.

When \(n\) is large \[ \frac{n^2}{\sqrt[3]{n^9+2}} \approx \frac{7n^2}{n^3} = \frac{7}{n} \] and so we have reason to believe the series diverges. Now \[ \frac{n^2}{\sqrt[3]{n^9+2}} \geq \frac{n^2}{\sqrt[3]{n^9+7n^9}} = \frac{n^2}{\sqrt[3]{8n^9}}=\frac{1}{2n} \] Since \(\sum \frac{1}{2n}\) diverges then the given series diverges. Alternatively, could use limit comparison test with \(b_n=\frac{1}{n}\).

Have negative terms so consider \[ \sumo |a_n| = \sumo \frac{3n+4}{7n^3+3n-1} \] Then with \(b_n=\frac{1}{n^2}\) we obtain \[ \limi \frac{|a_n|}{b_n} = \limi \frac{3n^3+4n^2}{7n^3+3n-1} = \frac{3}{7} \lt 1 \] Hence, by LCT, the series \(\sumo |a_n|\) converges because \(\sum b_n\) converges. Then by ACT, the original series \(\sum a_n\) converges absolutely.

Have negative terms so consider \[ \sumo |a_n| = \sumo \frac{|\cos(n)|}{n^2} \] Suspect converges but if we try the LCT with say \(b_n=\frac{1}{n^2}\) we get \[ \limi \frac{|a_n|}{b_n} = \limi |\cos(n)| \] and this limit DNE. Thus, the LCT is inconclusive. However, notice that \[ \frac{|\cos(n)|}{n^2} \leq \frac{1}{n^2} \] and \(\sum \frac{1}{n^2}\) converges and thus \(\sum |a_n|\) converges by the DCT. Thus, the given series \(\sum a_n\) converges absolutely.

Compute \[ \frac{a_{n+1}}{a_n} = \frac{10^{n+1}}{(n+1)!} \times \frac{n!}{10^n} = \frac{10}{n+1} \] and thus \[ \limi \left| \frac{a_{n+1}}{a_n} \right| = \limi \frac{10}{n+1}=0 = \rho \lt 1. \] Hence, the series converges by the RT.

Compute \begin{align*} \frac{a_{n+1}}{a_n} = \frac{(-1)^{n+1} (n+1)!}{e^{n+1}} \frac{e^n}{(-1)^n n!}=\frac{(-1) (n+1)}{e} \end{align*} and thus \[ \limi \left| \frac{a_{n+1}}{a_n} \right| = \limi \frac{(n+1)}{e} = \infty \] and thus by the RT the given series diverges.

We compute \begin{align*} \frac{a_{n+1}}{a_n} = \frac{4^{n+1}(n+1)!}{(2n+2)!} \frac{(2n)!}{4^n n!} = \frac{4 (n+1)}{(2n+2)(2n+1)} \end{align*} and therefore \[ \limi \left| \frac{a_{n+1}}{a_n} \right| = \limi \frac{4 (n+1)}{(2n+2)(2n+1)} = 0 \lt 1 \] and thus by the RT the given series converges.

We compute \begin{align*} \frac{a_{n+1}}{a_n} = \frac{(2n+2)!}{(n+1)!(n+1)!} \frac{n!n!}{(2n)!} = \frac{(2n+2)(2n+1)}{(n+1)(n+1)} \end{align*} and thus \[ \limi \left| \frac{a_{n+1}}{a_n} \right| = \limi \frac{(2n+2)(2n+1)}{(n+1)(n+1)} = 4 \gt 1 \] and thus by the RT the given series diverges.

Trying the ratio test gives \[ \limi \left| \frac{a_{n+1}}{a_n} \right| = 1 \] and the series of absolute values \(\sum |a_n| = \sum \frac{10n}{n^2+16}\) diverges, so the ACT is inconclusive. In this case, \[ u_n = \frac{10n}{n^2+16} \gt 0 \] and \(\limi u_n = 0\). Now \[ (u_n) = \left(\tfrac{10}{17}, \tfrac{20}{20}, \tfrac{30}{25}, \tfrac{40}{32}, \tfrac{50}{41}, \tfrac{60}{52}, \tfrac{70}{65}\ldots \right) \] we get that \(u_4 \lt u_5 \lt u_6 \lt \cdots\). To show this, let \(f(x) = \frac{10x}{x^2+16}\). Then \[ f'(x) = \frac{10(x^2+16)-20x^2}{(x^2+16)^2}=\frac{160-10x^2}{(x^2+16)^2} = \frac{10(16-x^2)}{(x^2+16)^2} \] Now \(f'(x) \lt 0\) if and only if \(x \gt 4\) and \(x \lt -4\). Hence, \((u_n)\) is decreasing starting with \(n\geq 4\). 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, \(u_n = \frac{n^2}{n^3+1} \lt 0\) and clearly \(\limi u_n = 0\). Now let \(f(x) = \frac{x^2}{x^3+1}\). Then \[ f'(x) = \frac{x(2-x^3}{(x^3+1)^2} \lt 0 \] iff \(x\geq 2^{1/3}\). So, \(u_n\) decreasing for \(n\geq 2\). Thus, the series converges conditionally.

Here the center is \(a=1\). Suppose we set \(x=3\): \[ \sumo \frac{(-1)^n 2^n}{n3^n} \] Apply the Ratio test: \begin{align*} \limi \left| \frac{a_{n+1}}{a_n} \right| &= \frac{2}{3} \lt 1 \end{align*} Thus, when \(x=3\), the resulting series converges. If instead we set \(x=-3\): \[ \sumo \frac{n(-1)^n (-4)^n}{3^n} \] and in this case the Ratio test gives \[ \limi \left| \frac{a_{n+1}}{a_n} \right| = \frac{4}{3} \gt 1 \] and thus the resulting series diverges when \(x=-3\). Apply the Ratio test for general \(x\): \begin{align*} \limi \left| \frac{a_{n+1}}{a_n}\right| &= \limi \left| \frac{(-1)^{n+1}(x-1)^{n+1}}{(n+1)3^{n+1}} \cdot \frac{3^nn}{(-1)^n (x-1)^n}\right|\\ &= |x-1| \limi \frac{n}{3(n+1)}\\ &= |x-1| \cdot\frac{1}{3} = \rho \end{align*} By the Ratio test, the series converges when \(\rho \lt 1\) or when \(|x-1| \lt 3\), or when \(x \in (1-3, 1 +3)=(-2,4)\). Note that the R.T. is inconclusive when \(\rho =1\) and when \(\rho \gt 1\) the series diverges. The case \(\rho=1\) occurs when \(|x-1| = 3\) which happens at the boundary points \(x=-2\) and \(x=4\). We need to check these manually. At \(x=-2\) we obtain \[ \sumo \frac{(-1)^n (-3)^n}{n3^n} = \sumo \frac{1}{n} \] which diverges. On the other hand, at the boundary point \(x=4\) we obtain \[ \sumo \frac{(-1)^n 3^n}{n3^n} = \sumo \frac{(-1)^n}{n} \] which converges. Hence, the interval of convergence is \((-2,4]\). We can therefore define the function \[ f(x) = \sumo \frac{(-1)^n}{n3^n}(x-1)^n \] whose domain is \((-2,4]\).

Applying the Ratio test yields \[ \limi \left|\frac{a_{n+1}}{a_n}\right| = \frac{|x+2|}{10} = \rho \] Hence, by the Ratio test, the series converges if \(|x+2| \lt 10\) or when \(x\in (-12,8)\). When \(\rho=1\), the Ratio test is inconclusive, which occurs when \(x=-12\) or \(x=8\). At \(x=-12\) we obtain \(\sumz (-1)^n\) which diverges and at \(x=8\) we obtain \(\sumz 1\) which also diverges. Hence, the interval where the series converges is \((-12,8)\).

We know that \[ \frac{1}{1-u} = \sumz u^n \] and this is valid for \(|u| \lt 1\) or for \(u\in (-1,1)\). Notice that \begin{align*} f(x) = \frac{8}{5} \left(\frac{1}{1+\tfrac{2x}{5}}\right) &= \frac{8}{5}\left(\frac{1}{1-(-\tfrac{2x}{5})}\right)\\[2ex] &=\frac{8}{5} \sumz (-\tfrac{2x}{5})^n\\ &= \sumz \frac{8(-1)^n 2^n x^n}{5^{n+1}} \end{align*} and this is valid for \(|\tfrac{-2x}{5}| \lt 1\) or for \(|x| \lt \frac{5}{2}\).

Notice that \[ f(x) = \sumz (-(2x+1))^n \] This is a geometric series with \(r=-(2x+1)\) and we know a geometric series converges when \(|-(2x+1)| \lt 1\): \[ |-(2x+1)| \lt 1 \Longrightarrow 2|(x+\tfrac{1}{2})| \lt 1 \Longrightarrow |x+\tfrac{1}{2}| \lt \tfrac{1}{2} \] Hence, interval of convergence is \((-1/2 - 1/2, -1/2 + 1/2) =(-1,0)\) and the closed-form expression is \[ f(x) = \frac{1}{1-(-(2x+1))} = \frac{1}{1+2x+1} = \frac{1}{2x+2} \]

Notice that \begin{align*} f(x) = x\left(\frac{1}{1-(-3x^2)}\right) &= x \sumz (-3x^2)^n \\[2ex] &= \sumz (-1)^n 3^n x^{2n+1} \end{align*} and the representation is valid when \(|-3x^2| \lt 1\): \[ |-3x^2| \lt 1 \Longrightarrow 3|x^2| \lt 1 \Longrightarrow |x| \lt \sqrt{1/3} \] Hence, the interval of convergence is \((-\sqrt{1/3},\sqrt{1/3})\). The power series we found is of the form \[ f(x) = \frac{x}{1+3x^2} = \sumz (-1)^n 3^n x^{2n+1} \] In our representation, the polynomial \(x^7\) appears when \(n=3\), hence \(c_7 = (-1)^3 3^3 = -27\) and since all powers of \(x\) appearing in the series are odd then \(c_8 = 0\).

We know that \[ f'(x) = \frac{1}{1+x^2} \] Now \[ \frac{1}{1+x^2} = \sumz (-x^2)^n = \sumz (-1)^n x^{2n} \] and this representation is valid when \(|-x^2| \lt 1\) or \(|x| \lt 1\). Therefore, \begin{align*} \arctan(x) = \int \frac{1}{1+x^2}\, dx &= \int \sumz (-1)^n x^{2n} \, dx \\[2ex] &= C + \sumz (-1)^n \frac{x^{2n+1}}{2n+1} \end{align*} To find \(C\) we evaluate both sides at a convenient point, say \(x=0\) and this gives \(C=0\). Hence, \[ \arctan(x) = \sumz (-1)^n \frac{x^{2n+1}}{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots \] and this is valid for \((-1,1)\). However, notice that at \(x=1\) we get \[ \sumz \frac{(-1)^n}{2n+1} \] which by the Alternating series test converges. Hence, \[ \arctan(1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots \] and since \(\arctan(1) = \frac{\pi}{4}\) we get \[ \frac{\pi}{4} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots \]

If we can find a power series representation for \(\ln(2+3x)\) then we will need to multiply the series by \(x\) to obtain a power series for \(f(x)\). Now, \[ \frac{d}{dx} \ln(2+3x) = \frac{3}{2+3x} \] so we will find a power series for \(\frac{3}{2+3x}\) and then integrate to get a power series for \(\ln(2+3x)\). Now \begin{align*} \frac{3}{2+3x} = \frac{3/2}{1+3x/2} = \sumz \frac{3}{2} (-3x/2)^n = \sumz \frac{(-1)^n 3^{n+1} x^n}{2^{n+1}} \end{align*} valid for \(|-3x/2| \lt 1\) or for \(|x| \lt 2/3\). Then \[ \ln(2+3x) = \int \sumz \frac{(-1)^n 3^{n+1} x^n}{2^{n+1}} \, dx = C + \sumz \frac{(-1)^n 3^{n+1} x^{n+1}}{2^{n+1}(n+1)} \] and valid for \(|x| \lt 2/3\). Now, evaluating both sides at \(x=0\) gives \(C=\ln(2)\). Hence \[ x\ln(2+3x) = x \ln(2) + \sumz \frac{(-1)^n 3^{n+1} x^{n+2}}{2^{n+1}(n+1)} \]

We know that \[ \frac{1}{1+x} = \sumz (-x)^n = \sumz (-1)^n x^n \] valid in the interval \((-1,1)\). Now, \[ \frac{d}{dx} \frac{1}{1+x} = \frac{-1}{(1+x)^2} \] Therefore, \[ \frac{-1}{(1+x)^2} = \frac{d}{dx} \sumz (-1)^n x^n = \sumo (-1)^n n x^{n-1} \] and thus \begin{align*} \frac{1}{(1+x)^2} &= \sumo (-1)^{n+1} n x^{n-1} \\[2ex] &= 1 - 2x + 3x^2 - 4x^3 + \cdots \end{align*} and the representation is valid in the interval \((-1,1)\).

Compute directly \begin{align*} f(x) &= \int \frac{1}{1+x^7}\, dx \\[2ex] &= \int \sumz (-x^7)^n \, dx\\[2ex] &= C + \sumz (-1)^n \frac{x^{7n+1}}{7n+1} \end{align*} Since \(f(0)=-1\) then \(C=0\) and the interval is \((-1,1)\). The power \(x^3\) does not appear and thus \(c_3=0\). In our representation, the term \(x^{15}\) corresponds to \(n=3\) and thus \(c_{22} = \frac{-1}{22}\).

We compute \begin{align*} f(x) &= \sin(x) & f(0) &= 0 \\ f'(x) &= \cos(x) & f'(0) &= 1 \\ f^{(2)}(x) &= -\sin(x) & f^{(2)}(0) &= 0 \\ f^{(3)}(x) &= -\cos(x) & f^{(3)}(0) &= -1\\ f^{(4)}(x) &= \sin(x) & f^{(4)}(0) &= 0 \\ f^{(5)}(x) &= \cos(x) & f^{(5)}(0) &= 1 \end{align*} and the pattern repeats. We get \begin{align*} \sumz \frac{f^{(n)}(0)}{n!} x^n &= f(0) + f'(0) x + \frac{f^{(2)}(0)}{2!} x^2 + \cdots \\ &= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \frac{x^9}{9!} - \cdots\\ &= \sumz (-1)^n \frac{x^{2n+1}}{(2n+1)!} \end{align*} Therefore, \[ T_8(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} \] Performing the Ratio test on the Maclaurin series we get \[ \limi \left|\frac{a_{n+1}}{a_n}\right| = 0 \] and thus the interval of convergence is \((-\infty,\infty)\). Therefore, \[ \sin(x) = \sumz (-1)^n \frac{x^{2n+1}}{(2n+1)!} \]

We could proceed as was done for \(\sin(x)\) or we could use the fact that \(\frac{d}{dx} \sin(x) = \cos(x)\). Differentiating the power series of \(\sin(x)\) we obtain \begin{align*} \cos(x) &= \sumz (-1)^n \frac{x^{2n}}{(2n)!}\\ &= 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots \end{align*} with interval of convergence \((-\infty, \infty)\).

We need to compute \(f^{(n)}(0)\). But since \(f^{(n)}(x) = e^x\) for all \(n\) then \(f^{(n)}(0) = 1\). Thus the Maclaurin series is \[ \sumz \frac{f^{(n)}(0)}{n!} x^n = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + \cdots \] And thus \[ T_4(x) = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} \] To find the interval of convergence compute \[ \limi \left|\frac{a_{n+1}}{a_n}\right| = 0 \] and thus the interval of convergence is \((-\infty, \infty)\). Therefore, \[ e^x = \sumz \frac{x^n}{n!} \] This shows, for instance, that \[ e = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \frac{1}{4!} + \cdots \]

Compute \begin{align*} f(x) &= e^{2x} & f(1) &= e^2\\ f'(x) &= 2e^{2x} & f'(1) &= 2e^2\\ f^{(2)}(x) &= 2^2 e^{2x} & f^{(2)}(1) &= 2^2 e^2\\ f^{(3)}(x) &= 2^3 e^{2x} & f^{(3)}(1) &= 2^3 \end{align*} The pattern is \(f^{(n)}(1) = 2^n e^2\). Therefore, the series centered at \(x=1\) is \[ \sumz \frac{f^{(n)}(1)}{n!} (x-1)^n = \sumz \frac{2^n e^2}{n!} (x-1)^n \]

Compute \begin{align*} f(x) = \int_0^x e^{t^3}\,dx &= \int_0^x \sumz \frac{1}{n!} t^{3n}\\ &= \sumz \frac{1}{n!} \frac{t^{3n+1}}{(3n+1)}\Big|_0^x\\ &= \sumz \frac{1}{n!} \frac{x^{3n+1}}{(3n+1)} \end{align*} Because the power series of \(e^x\) converges on \((-\infty,\infty)\) then the Taylor series of \(f(x)\) converges on \((-\infty,\infty)\). To find \(f^{(10)}(0)\) recall that a function's power series is its Taylor series, so if \[ f(x) = \sumz c_n (x-a)^n \] then \[ \sumz c_n (x-a)^n = \sumz \frac{f^{(n)}(a)}{n!} (x-a)^n \] Then \[ c_n = \frac{f^{(n)}(a)}{n!} \] and thus \[ f^{(n)}(a) = n! c_n \] Now from our power series representation, the term \(x^{10}\) corresponds to \(n=3\) and thus \[ c_{10} = \frac{1}{3! 10} = \frac{1}{60} \] Then \[ f^{(10)}(0) = 10! \frac{1}{60} \] Now to find \(T_8(x)\) we expand and keep only up to the term \(x^8\) if present: \[ \sumz \frac{1}{n!} \frac{x^{3n+1}}{(3n+1)} = x + \frac{1}{1!\cdot 4} x^4 + \frac{1}{2!\cdot 7} x^7 + \frac{1}{3!\cdot 10} x^{10} + \cdots \] and therefore \[ T_8(x) = x + \frac{1}{1!\cdot 4} x^4 + \frac{1}{2!\cdot 7} x^7 \] Then \[ f(0.5) \approx T_8(0.5) = \frac{1}{2} + \frac{1}{4\cdot 2^4} + \frac{1}{14\cdot 2^7} = \frac{925}{1792} = 0.516183\ldots \]

We have \begin{align*} f(x) &= \frac{1}{3}\left(2x + x\sumz \frac{(-1)^n (\tfrac{x}{5})^{2n+1}}{(2n+1)!}\right) \\[2ex] &= \frac{2}{3}x + \frac{x}{3} \sumz \frac{(-1)^n x^{2n+1}}{5^{2n+1}(2n+1)!} \\[2ex] &= \frac{2}{3}x + \sumz \frac{(-1)^n x^{2n+2}}{3\cdot 5^{2n+1}(2n+1)!} \end{align*} Then \[ f^{(22)}(0) = \frac{(-1)^{10}}{3 \cdot 5^{21} (21)!} \]

After successive differentiation one finds that \[ f^{(n)}(x) = ne^x + xe^x \] and thus \(f^{(n)}(1) = (n+1)e\). Thus the Taylor series is \[ \sumz \frac{(n+1)e}{n!} (x-1)^n \] and then \[ T_3(x) = e + 2e(x-1) + \frac{3e}{2!}(x-1)^2 + \frac{4e}{4!}(x-1)^3 \]