Number Series and Function Series: Problems with Solutions

Solution:
For [tex]-1\le x\le 1[/tex] the series is geometric with first term [tex]1[/tex] and ratio [tex]q=\frac{x}{2}[/tex], where [tex]|q|\le\frac{1}{2}\lt 1[/tex]. Hence it converges and
[tex]f(x)=\frac{1}{1-\frac{x}{2}}=\frac{2}{2-x}[/tex]
Then
[tex]\int_{0}^{1}\frac{2}{2-x}\,dx=\left[-2\ln(2-x)\right]_{0}^{1}=-2\ln 1+2\ln 2=2\ln 2[/tex]
Hence
[tex]\sqrt{e^{2\ln 2}}=\sqrt{e^{\ln 4}}=\sqrt{4}=2[/tex]

Solution:
Since [tex]1-\frac{1}{n}=\frac{n-1}{n}[/tex] and [tex]1-\frac{1}{n+1}=\frac{n}{n+1}[/tex], the general term is a difference of consecutive terms of the sequence [tex]b_n=\log_{2}\frac{n-1}{n}[/tex]:
[tex]a_n=\log_{2}\frac{n-1}{n}-\log_{2}\frac{n}{n+1}=b_n-b_{n+1}[/tex]
Therefore the partial sums telescope:
[tex]S_N=\sum_{n=2}^{N}(b_n-b_{n+1})=b_2-b_{N+1}=\log_{2}\frac{1}{2}-\log_{2}\frac{N}{N+1}[/tex]
As [tex]N\to\infty[/tex] we have [tex]\frac{N}{N+1}\to 1[/tex], so [tex]\log_{2}\frac{N}{N+1}\to 0[/tex]. Hence the series converges and its sum is
[tex]S=\lim_{N\to\infty}S_N=\log_{2}\frac{1}{2}=-1[/tex]

Solution:
Decompose the general term into partial fractions (note that [tex](2n+5)-(2n-1)=6[/tex]):
[tex]\frac{1}{(2n-1)(2n+5)}=\frac{1}{6}\left(\frac{1}{2n-1}-\frac{1}{2n+5}\right)\quad\Rightarrow\quad S_n=\frac{1}{6}\left(\sum_{k=1}^{n}\frac{1}{2k-1}-\sum_{k=1}^{n}\frac{1}{2k+5}\right)[/tex]
Since [tex]2k+5=2(k+3)-1[/tex], the second sum consists of the fractions [tex]\frac{1}{7},\frac{1}{9},\ldots,\frac{1}{2n+5}[/tex], i.e. of the same kind of fractions as the first sum, with the index shifted by [tex]3[/tex]. For [tex]n\ge 3[/tex] the fractions [tex]\frac{1}{7},\ldots,\frac{1}{2n-1}[/tex] cancel and
[tex]S_n=\frac{1}{6}\left(1+\frac{1}{3}+\frac{1}{5}-\frac{1}{2n+1}-\frac{1}{2n+3}-\frac{1}{2n+5}\right)[/tex]
As [tex]n\to\infty[/tex] the last three fractions tend to [tex]0[/tex], so the series converges and
[tex]S=\lim_{n\to\infty}S_n=\frac{1}{6}\times \frac{23}{15}=\frac{23}{90}[/tex]

Solution:
Rewrite the numerator of the general term:
[tex](-4)^{n}x^{6n}=(-1)^{n}2^{2n}\left(x^{3}\right)^{2n}=(-1)^{n}\left(2x^{3}\right)^{2n}[/tex]
Hence, with [tex]t=2x^{3}[/tex], the series becomes
[tex]\sum_{n=0}^{\infty}\frac{(-1)^{n}t^{2n}}{(2n)!}[/tex]
This is the Maclaurin series of [tex]\cos t[/tex]. It converges to [tex]\cos t[/tex] for every real [tex]t[/tex]: the radius of convergence is infinite, and the Lagrange remainder tends to [tex]0[/tex] because all derivatives of the cosine are bounded in absolute value by [tex]1[/tex]. Therefore, for every [tex]x\in\mathbb{R}[/tex] the sum of the series is
[tex]\cos\left(2x^{3}\right)[/tex]

Solution:
The series [tex]T=\sum_{n=1}^{\infty}\frac{n}{2^{n}}[/tex] and [tex]G=\sum_{n=1}^{\infty}\frac{1}{2^{n}}[/tex] converge (the first by the ratio test, since [tex]\frac{n+1}{2n}\to\frac{1}{2}\lt 1[/tex]; the second is geometric). Hence the given series converges too, and its sum is [tex]S=2T-G[/tex], where
[tex]G=\frac{\frac{1}{2}}{1-\frac{1}{2}}=1[/tex]
To find [tex]T[/tex], shift the index in [tex]\frac{T}{2}[/tex] and subtract:
[tex]\frac{T}{2}=\sum_{n=1}^{\infty}\frac{n}{2^{n+1}}=\sum_{n=2}^{\infty}\frac{n-1}{2^{n}}[/tex]
[tex]T-\frac{T}{2}=\frac{1}{2}+\sum_{n=2}^{\infty}\frac{n-(n-1)}{2^{n}}=\frac{1}{2}+\frac{1}{2}=1\quad\Rightarrow\quad T=2[/tex]
Therefore
[tex]S=2\times 2-1=3[/tex]

Solution:
Apply the ratio test to the absolute values of the terms:
[tex]\left|\frac{a_{n+1}}{a_n}\right|=\frac{n+2}{n+3}\,|x-4|\to|x-4|[/tex]
Hence the series converges absolutely for [tex]|x-4|\lt 1[/tex], i.e. for [tex]3\lt x\lt 5[/tex], and diverges for [tex]|x-4|\gt 1[/tex]; the radius of convergence is [tex]R=1[/tex]. The endpoints must be checked separately. At [tex]x=5[/tex] we get a tail of the harmonic series, which diverges:
[tex]\sum_{n=1}^{\infty}\frac{1}{n+2}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\ldots[/tex]
At [tex]x=3[/tex] we have [tex](x-4)^{n+2}=(-1)^{n+2}=(-1)^{n}[/tex], and the series becomes
[tex]\sum_{n=1}^{\infty}\frac{(-1)^{n}}{n+2}[/tex]
The numbers [tex]\frac{1}{n+2}[/tex] decrease and tend to [tex]0[/tex], so by the Leibniz test this series converges (conditionally). Therefore the interval of convergence is [tex]\left[3 , 5\right)[/tex].

Solution:
Removing finitely many terms does not change whether a series converges, so it is enough to study a tail of the series. By the definition of the limit with [tex]\varepsilon=\frac{L}{2}\gt 0[/tex], there is an index [tex]N[/tex] such that for all [tex]n\ge N[/tex]
[tex]n\,a_n\gt L-\frac{L}{2}=\frac{L}{2}\quad\Rightarrow\quad a_n\gt\frac{L}{2}\times \frac{1}{n}\gt 0[/tex]
The harmonic series [tex]\sum_{n=1}^{\infty}\frac{1}{n}[/tex] diverges, hence so does [tex]\sum_{n=N}^{\infty}\frac{L}{2n}[/tex] (dropping finitely many terms and multiplying by the positive constant [tex]\frac{L}{2}[/tex] do not change divergence).
By the comparison test for series with positive terms, [tex]\sum_{n=N}^{\infty}a_n[/tex] diverges. Adding the finitely many terms [tex]a_1,\ldots,a_{N-1}[/tex] does not change this, so the series [tex]\sum_{n=1}^{\infty}a_n[/tex] diverges.

Solution:
The Fourier coefficients may be computed over any interval of length [tex]2\pi[/tex]; we take [tex]\left(0, 2\pi\right)[/tex]. First,
[tex]a_0=\frac{1}{\pi}\int_{0}^{2\pi}x\,dx=\frac{1}{\pi}\times \frac{(2\pi)^2}{2}=2\pi[/tex]
Hence the constant term is [tex]\frac{a_0}{2}=\pi[/tex]. Integrating by parts, for [tex]n\ge 1[/tex] we get
[tex]a_n=\frac{1}{\pi}\int_{0}^{2\pi}x\cos nx\,dx=\frac{1}{\pi}\left(\left[\frac{x\sin nx}{n}\right]_{0}^{2\pi}-\frac{1}{n}\int_{0}^{2\pi}\sin nx\,dx\right)=0[/tex]
[tex]b_n=\frac{1}{\pi}\int_{0}^{2\pi}x\sin nx\,dx=\frac{1}{\pi}\left(\left[-\frac{x\cos nx}{n}\right]_{0}^{2\pi}+\frac{1}{n}\int_{0}^{2\pi}\cos nx\,dx\right)=\frac{1}{\pi}\times \left(-\frac{2\pi}{n}\right)=-\frac{2}{n}[/tex]
Therefore
[tex]f(x)=\pi-2\sum_{n=1}^{\infty}\frac{\sin nx}{n}[/tex]
Since [tex]f[/tex] is piecewise smooth, by Dirichlet's theorem the series converges to [tex]x[/tex] for [tex]0\lt x\lt 2\pi[/tex]; at the jump points [tex]x=2k\pi,\ k\in\mathbb{Z}[/tex], its sum is [tex]\pi[/tex], the mean of the one-sided limits [tex]0[/tex] and [tex]2\pi[/tex].