Number Series and Function Series: Difficult Problems with Solutions

Solution:
Put [tex]t=-\frac{x^2}{5}[/tex]. The series becomes [tex]\sum_{n=0}^{\infty}(n+1)t^n[/tex], and [tex]|t|\lt 1[/tex] exactly when [tex]|x|\lt\sqrt{5}[/tex]. For [tex]|t|\lt 1[/tex] the geometric series gives
[tex]\sum_{n=0}^{\infty}t^{n+1}=\frac{t}{1-t}[/tex]
A power series may be differentiated term by term inside its interval of convergence, so
[tex]\sum_{n=0}^{\infty}(n+1)t^n=\left(\frac{t}{1-t}\right)'=\frac{(1-t)+t}{(1-t)^2}=\frac{1}{(1-t)^2}[/tex]
Substituting back [tex]t=-\frac{x^2}{5}[/tex], we obtain the sum:
[tex]S(x)=\frac{1}{\left(1+\frac{x^2}{5}\right)^2}=\frac{25}{\left(5+x^2\right)^2}[/tex]
Check: [tex]S(0)=1[/tex], which is the first term of the series.

Solution:
Instead of computing four derivatives, we compose known expansions. Let [tex]u=\sin x=x-\frac{x^3}{6}+o(x^4)[/tex]. Since [tex]u=O(x)[/tex] as [tex]x\to 0[/tex], we have
[tex]e^u=1+u+\frac{u^2}{2}+\frac{u^3}{6}+\frac{u^4}{24}+o(x^4)[/tex]
Up to [tex]o(x^4)[/tex]:
[tex]u^2=x^2-\frac{x^4}{3}+o(x^4), \quad u^3=x^3+o(x^4), \quad u^4=x^4+o(x^4)[/tex]
Substituting into the expansion of [tex]e^u[/tex]:
[tex]e^{\sin x}=1+\left(x-\frac{x^3}{6}\right)+\left(\frac{x^2}{2}-\frac{x^4}{6}\right)+\frac{x^3}{6}+\frac{x^4}{24}+o(x^4)[/tex]
The [tex]x^3[/tex] terms cancel and [tex]-\frac{1}{6}+\frac{1}{24}=-\frac{1}{8}[/tex], hence
[tex]e^{\sin x}=1+x+\frac{x^2}{2}-\frac{x^4}{8}+o(x^4)[/tex]
Check with derivatives: [tex]f(0)=1, \ f'(0)=1, \ f''(0)=1, \ f'''(0)=0, \ f^{(4)}(0)=-3[/tex], and [tex]\frac{-3}{4!}=-\frac{1}{8}[/tex].

Solution:
Extend [tex]f[/tex] to an even function on [tex]\left(-\pi, \pi\right)[/tex] and then periodically with period [tex]2\pi[/tex]. Then [tex]b_n=0[/tex] and
[tex]a_n=\frac{2}{\pi}\int_{0}^{\pi}f(x)\cos nx\,dx[/tex]
First,
[tex]a_0=\frac{2}{\pi}\int_{0}^{\pi}\left(\frac{\pi}{4}-\frac{x}{2}\right)dx=\frac{2}{\pi}\left(\frac{\pi^{2}}{4}-\frac{\pi^{2}}{4}\right)=0[/tex]
For [tex]n\ge 1[/tex] we have [tex]\int_{0}^{\pi}\cos nx\,dx=0[/tex] and, integrating by parts,
[tex]\int_{0}^{\pi}x\cos nx\,dx=\left[\frac{x\sin nx}{n}\right]_{0}^{\pi}+\left[\frac{\cos nx}{n^{2}}\right]_{0}^{\pi}=\frac{(-1)^{n}-1}{n^{2}}[/tex]
Hence
[tex]a_n=\frac{2}{\pi}\times \left(-\frac{1}{2}\right)\times \frac{(-1)^{n}-1}{n^{2}}=\frac{1-(-1)^{n}}{\pi n^{2}}[/tex]
This equals [tex]0[/tex] for even [tex]n[/tex] and [tex]\frac{2}{\pi n^{2}}[/tex] for odd [tex]n=2k-1[/tex]. The even extension is continuous and piecewise smooth, so by Dirichlet's theorem the series converges to [tex]\frac{\pi}{4}-\frac{x}{2}[/tex] on the whole segment [tex]\left[0, \pi\right][/tex]:
[tex]\frac{\pi}{4}-\frac{x}{2}=\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{\cos\left((2k-1)x\right)}{(2k-1)^{2}}[/tex]
Check at [tex]x=0[/tex]: since [tex]\sum_{k=1}^{\infty}\frac{1}{(2k-1)^{2}}=\frac{\pi^{2}}{8}[/tex], the right-hand side equals [tex]\frac{2}{\pi}\times \frac{\pi^{2}}{8}=\frac{\pi}{4}[/tex].

Solution:
We expand [tex]f[/tex] in a Maclaurin series using the geometric series. For [tex]\left|\frac{5}{3}x^{15}\right|\lt 1[/tex]:
[tex]\frac{1}{3+5x^{15}}=\frac{1}{3}\times \frac{1}{1+\frac{5}{3}x^{15}}=\frac{1}{3}\sum_{m=0}^{\infty}\left(-\frac{5}{3}\right)^m x^{15m}[/tex]
Since [tex]\frac{2+3x^{12}}{3}=\frac{2}{3}+x^{12}[/tex], we get
[tex]f(x)=\sum_{m=0}^{\infty}\frac{2}{3}\left(-\frac{5}{3}\right)^m x^{15m}+\sum_{m=0}^{\infty}\left(-\frac{5}{3}\right)^m x^{15m+12}[/tex]
A convergent power series is the Taylor series of its sum, so [tex]f^{(k)}(0)=k!\,c_k[/tex], where [tex]c_k[/tex] is the coefficient of [tex]x^k[/tex]. Since [tex]87=15\times 5+12[/tex] and [tex]87[/tex] is not divisible by [tex]15[/tex], the power [tex]x^{87}[/tex] occurs only in the second sum, with [tex]m=5[/tex]:
[tex]c_{87}=\left(-\frac{5}{3}\right)^5=-\frac{3125}{243}[/tex]
Hence
[tex]f^{(87)}(0)=87!\times c_{87}=-\frac{3125}{243}\times 87![/tex]

Solution:
Divide the polynomials: [tex]n^2+1=(n+2)(n-2)+5[/tex], so
[tex]\frac{n^2+1}{n+2}=n-2+\frac{5}{n+2}[/tex]
Each of the series below converges (for example, by the ratio test), so the given series may be split:
[tex]S=\sum_{n=1}^{\infty}\frac{n}{n!}-2\sum_{n=1}^{\infty}\frac{1}{n!}+5\sum_{n=1}^{\infty}\frac{1}{(n+2)\,n!}[/tex]
We use the Maclaurin series [tex]e=\sum_{n=0}^{\infty}\frac{1}{n!}[/tex]. In the second sum the term [tex]\frac{1}{0!}=1[/tex] is missing:
[tex]\sum_{n=1}^{\infty}\frac{n}{n!}=\sum_{n=1}^{\infty}\frac{1}{(n-1)!}=e, \qquad \sum_{n=1}^{\infty}\frac{1}{n!}=e-1[/tex]
For the third sum,
[tex]\frac{1}{(n+2)\,n!}=\frac{n+1}{(n+2)!}=\frac{(n+2)-1}{(n+2)!}=\frac{1}{(n+1)!}-\frac{1}{(n+2)!}[/tex]
so the sum telescopes:
[tex]\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)!}-\frac{1}{(n+2)!}\right)=\frac{1}{2!}=\frac{1}{2}[/tex]
Hence
[tex]S=e-2(e-1)+\frac{5}{2}=\frac{9}{2}-e[/tex]

Solution:
By the ratio test,
[tex]\left|\frac{x^{n+3}}{(n+2)(n+3)}\times \frac{(n+1)(n+2)}{x^{n+2}}\right|=\frac{n+1}{n+3}\,|x|\to|x|[/tex]
Hence the radius of convergence is [tex]R=1[/tex], and inside [tex]\left(-1, 1\right)[/tex] the series may be differentiated term by term. Let [tex]S(x)[/tex] be its sum; then [tex]S(0)=0[/tex], [tex]S'(0)=0[/tex], and the second derivative is a geometric series:
[tex]S'(x)=\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}, \qquad S''(x)=\sum_{n=0}^{\infty}x^{n}=\frac{1}{1-x}[/tex]
Integrating from [tex]0[/tex] to [tex]x[/tex] and using [tex]S'(0)=0[/tex]:
[tex]S'(x)=\int_{0}^{x}\frac{dt}{1-t}=-\ln(1-x)[/tex]
We integrate again, using [tex]S(0)=0[/tex]. With the substitution [tex]u=1-t[/tex],
[tex]\int_{0}^{x}\ln(1-t)\,dt=\int_{1-x}^{1}\ln u\,du=\left[u\ln u-u\right]_{1-x}^{1}=-x-(1-x)\ln(1-x)[/tex]
Therefore
[tex]S(x)=-\int_{0}^{x}\ln(1-t)\,dt=x+(1-x)\ln(1-x)[/tex]
Check: the Maclaurin expansion of the right-hand side begins with [tex]\frac{x^{2}}{2}+\frac{x^{3}}{6}+\frac{x^{4}}{12}+\ldots[/tex], which matches the series.

Solution:
We look for a telescoping decomposition. Put [tex]b_n=\frac{2^n}{3^n-2^n}[/tex]. Then
[tex]b_n-b_{n+1}=\frac{2^n\left(3^{n+1}-2^{n+1}\right)-2^{n+1}\left(3^n-2^n\right)}{\left(3^{n+1}-2^{n+1}\right)\left(3^n-2^n\right)}[/tex]
and the numerator equals
[tex]3\times 6^n-2^{2n+1}-2\times 6^n+2^{2n+1}=6^n[/tex]
So the general term of the series is exactly [tex]b_n-b_{n+1}[/tex], and the partial sums telescope:
[tex]S_N=b_1-b_{N+1}=\frac{2}{3-2}-\frac{2^{N+1}}{3^{N+1}-2^{N+1}}=2-\frac{1}{\left(\frac{3}{2}\right)^{N+1}-1}[/tex]
Since [tex]\left(\frac{3}{2}\right)^{N+1}\to\infty[/tex], the last fraction tends to [tex]0[/tex]. Hence the series converges and its sum is
[tex]\lim_{N\to\infty}S_N=2[/tex]

Solution:
We use the theorem on term-by-term differentiation: if the functions [tex]u_n[/tex] are continuously differentiable on a segment [tex]\left[a, b\right][/tex], the series [tex]\sum u_n(x_0)[/tex] converges at some point [tex]x_0\in\left[a, b\right][/tex] and the series of derivatives [tex]\sum u_n'(x)[/tex] converges uniformly on [tex]\left[a, b\right][/tex], then [tex]\sum u_n[/tex] converges uniformly on [tex]\left[a, b\right][/tex], its sum is differentiable there and [tex]\left(\sum u_n\right)'=\sum u_n'[/tex].
Here the functions [tex]u_n(x)=\frac{\sin \left(n^{3}x\right)}{2n^{5}+3}[/tex] are continuously differentiable on [tex]\mathbb{R}[/tex] and
[tex]u_n'(x)=\frac{n^{3}\cos\left(n^{3}x\right)}{2n^{5}+3}[/tex]
At [tex]x_0=0[/tex] all terms are equal to [tex]0[/tex], so the series converges there. For every real [tex]x[/tex]
[tex]\left|u_n'(x)\right|\le\frac{n^{3}}{2n^{5}+3}\lt\frac{n^{3}}{2n^{5}}=\frac{1}{2n^{2}}[/tex]
and the series [tex]\sum\frac{1}{2n^{2}}[/tex] converges (generalized harmonic series with [tex]p=2\gt 1[/tex]). By the Weierstrass M-test the series of derivatives converges uniformly on [tex]\mathbb{R}[/tex], hence on every segment [tex]\left[a, b\right][/tex].
Applying the theorem on an arbitrary segment containing a given point [tex]x[/tex], we get that [tex]S[/tex] is differentiable at [tex]x[/tex] and [tex]S'(x)=\sum_{n=1}^{\infty}\frac{n^{3}\cos\left(n^{3}x\right)}{2n^{5}+3}[/tex]. Since [tex]x\in\mathbb{R}[/tex] is arbitrary, the statement is proved (moreover, [tex]S'[/tex] is continuous as a uniform limit of continuous functions).

Solution:
Denote the general term by [tex]u_n[/tex] and apply the Cauchy root test to [tex]\left|u_n\right|[/tex]:
[tex]\sqrt[n]{\left|u_n\right|}=\frac{1}{\pi}\left|a+\frac{1}{n}\right|^{n}[/tex]
Case [tex]|a|\lt 1[/tex]: for large [tex]n[/tex] we have [tex]\left|a+\frac{1}{n}\right|\le q=\frac{1+|a|}{2}\lt 1[/tex], so [tex]\sqrt[n]{\left|u_n\right|}\le\frac{q^n}{\pi}\to 0\lt 1[/tex] and the series converges absolutely.
Case [tex]a=1[/tex]: [tex]\sqrt[n]{u_n}=\frac{1}{\pi}\left(1+\frac{1}{n}\right)^{n}\to\frac{e}{\pi}\lt 1[/tex], so the series converges.
Case [tex]a=-1[/tex]: [tex]\sqrt[n]{\left|u_n\right|}=\frac{1}{\pi}\left(1-\frac{1}{n}\right)^{n}\to\frac{1}{e\pi}\lt 1[/tex], so the series converges absolutely.
Case [tex]|a|\gt 1[/tex]: for large [tex]n[/tex] we have [tex]\left|a+\frac{1}{n}\right|\ge|a|-\frac{1}{n}\ge r=\frac{1+|a|}{2}\gt 1[/tex], so
[tex]\left|u_n\right|\ge\left(\frac{r^{n}}{\pi}\right)^{n}\to+\infty[/tex]
The general term does not tend to [tex]0[/tex], so the series diverges. Hence the series converges exactly for [tex]a\in\left[-1, 1\right][/tex].

Solution:
The function [tex]x^{2}[/tex] is even, so all sine coefficients vanish, [tex]b_n=0[/tex], and
[tex]a_n=\frac{2}{\pi}\int_{0}^{\pi}x^{2}\cos nx\,dx[/tex]
For [tex]n=0[/tex] we get [tex]a_0=\frac{2}{\pi}\times \frac{\pi^{3}}{3}=\frac{2\pi^{2}}{3}[/tex]. For [tex]n\ge 1[/tex] we integrate by parts twice:
[tex]\int_{0}^{\pi}x^{2}\cos nx\,dx=\left[\frac{x^{2}\sin nx}{n}\right]_{0}^{\pi}-\frac{2}{n}\int_{0}^{\pi}x\sin nx\,dx=-\frac{2}{n}\int_{0}^{\pi}x\sin nx\,dx[/tex]
[tex]\int_{0}^{\pi}x\sin nx\,dx=\left[-\frac{x\cos nx}{n}\right]_{0}^{\pi}+\frac{1}{n}\int_{0}^{\pi}\cos nx\,dx=-\frac{\pi(-1)^{n}}{n}[/tex]
Therefore
[tex]\int_{0}^{\pi}x^{2}\cos nx\,dx=\frac{2\pi(-1)^{n}}{n^{2}},\qquad a_n=\frac{4(-1)^{n}}{n^{2}}[/tex]
The [tex]2\pi[/tex]-periodic extension of [tex]x^{2}[/tex] from [tex]\left[-\pi, \pi\right][/tex] is continuous (its values at [tex]\pm\pi[/tex] coincide) and piecewise smooth, so by Dirichlet's theorem its Fourier series converges to it at every point:
[tex]x^{2}=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos nx=\frac{\pi^{2}}{3}+4\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n^{2}}\cos nx,\quad x\in\left[-\pi, \pi\right][/tex]
Put [tex]x=\pi[/tex]. Since [tex]\cos n\pi=(-1)^{n}[/tex], we have [tex](-1)^{n}\cos n\pi=1[/tex] and
[tex]\pi^{2}=\frac{\pi^{2}}{3}+4\sum_{n=1}^{\infty}\frac{1}{n^{2}}\Rightarrow\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{1}{4}\left(\pi^{2}-\frac{\pi^{2}}{3}\right)=\frac{\pi^{2}}{6}[/tex]

Solution:
The ratio test cannot be applied directly, because [tex]\frac{\cos(k+1)}{\cos k}[/tex] has no limit, so we estimate the terms.
Let [tex]|x|\lt 3[/tex] and [tex]q=\frac{|x|}{3}\lt 1[/tex]. Then
[tex]\left|\frac{k^{3}\cos k}{3^{k}}x^{k}\right|\le k^{3}q^{k},\qquad\frac{(k+1)^{3}q^{k+1}}{k^{3}q^{k}}\to q\lt 1[/tex]
so [tex]\sum k^{3}q^{k}[/tex] converges by the ratio test, and by comparison the power series converges absolutely.
For every [tex]k[/tex] at least one of the numbers [tex]|\cos k|[/tex] and [tex]|\cos(k+1)|[/tex] is at least [tex]\frac{1}{4}[/tex]. Indeed, if both were smaller than [tex]\frac{1}{4}[/tex], then from [tex]\cos(k+1)=\cos k\cos 1-\sin k\sin 1[/tex] we would get
[tex]|\sin k|\sin 1\le|\cos k|\cos 1+|\cos(k+1)|\lt\frac{1}{2}\Rightarrow|\sin k|\lt\frac{1}{2\sin 1}\lt 0.6[/tex]
(since [tex]\sin 1\gt 0.84[/tex]), while [tex]|\sin k|=\sqrt{1-\cos^{2}k}\gt\sqrt{\frac{15}{16}}\gt 0.96[/tex], a contradiction. Hence [tex]|\cos k|\ge\frac{1}{4}[/tex] for infinitely many [tex]k[/tex].
Now let [tex]|x|\ge 3[/tex]. For these infinitely many [tex]k[/tex]
[tex]\left|\frac{k^{3}\cos k}{3^{k}}x^{k}\right|\ge\frac{k^{3}}{4}\left(\frac{|x|}{3}\right)^{k}\ge\frac{k^{3}}{4}[/tex]
so the terms do not tend to [tex]0[/tex] and the series diverges. Therefore the series converges for [tex]|x|\lt 3[/tex] and diverges for [tex]|x|\ge 3[/tex], i.e. [tex]R=3[/tex].

Solution:
Put [tex]u_n=\frac{4^n(n!)^2}{(2n)!\,(n+1)}\gt 0[/tex]; the series is [tex]\sum(-1)^{n-1}u_n[/tex]. Since [tex]4^n(n!)^2=\left(2\times 4\times \ldots\times 2n\right)^2=\left((2n)!!\right)^2[/tex] and [tex](2n)!=(2n)!!\,(2n-1)!![/tex], we get
[tex]u_n=\frac{P_n}{n+1},\qquad P_n=\frac{(2n)!!}{(2n-1)!!}=\prod_{k=1}^{n}\frac{2k}{2k-1}[/tex]
Lower bound: from [tex](2k)^2\ge(2k-1)(2k+1)[/tex] we get [tex]\left(\frac{2k}{2k-1}\right)^2\ge\frac{2k+1}{2k-1}[/tex], and the product telescopes:
[tex]P_n^2\ge\prod_{k=1}^{n}\frac{2k+1}{2k-1}=2n+1[/tex]
Upper bound: for [tex]k\ge 2[/tex] the inequality [tex]4k(k-1)\le(2k-1)^2[/tex] gives [tex]\left(\frac{2k}{2k-1}\right)^2\le\frac{k}{k-1}[/tex], so for [tex]n\ge 1[/tex]
[tex]P_n^2\le 4\prod_{k=2}^{n}\frac{k}{k-1}=4n,\qquad\sqrt{2n+1}\le P_n\le 2\sqrt{n}[/tex]
No absolute convergence: for [tex]n\ge 1[/tex]
[tex]u_n\ge\frac{\sqrt{2n+1}}{n+1}\ge\frac{\sqrt{n}}{2n}=\frac{1}{2\sqrt{n}}[/tex]
and [tex]\sum\frac{1}{\sqrt{n}}[/tex] diverges ([tex]p[/tex]-series with [tex]p=\frac{1}{2}\le 1[/tex]), so [tex]\sum\left|(-1)^{n-1}u_n\right|=\sum u_n[/tex] diverges by comparison.
Convergence: [tex]u_n\le\frac{2\sqrt{n}}{n+1}\lt\frac{2}{\sqrt{n}}\to 0[/tex], and for all [tex]n\ge 0[/tex]
[tex]\frac{u_{n+1}}{u_n}=\frac{4(n+1)^2}{(2n+1)(2n+2)}\times \frac{n+1}{n+2}=\frac{2n^2+4n+2}{2n^2+5n+2}\le 1[/tex]
so [tex]u_n[/tex] is non-increasing and tends to [tex]0[/tex]. By the Leibniz test the alternating series converges. Being convergent but not absolutely convergent, the series is conditionally convergent.