Number Series and Function Series: Very Difficult Problems with Solutions

Solution:
Let [tex]a_n=\left(1-\frac{1}{n}\right)^{n^{2}}[/tex]. By the Cauchy-Hadamard formula
[tex]\sqrt[n]{a_n}=\left(1-\frac{1}{n}\right)^{n}\to\frac{1}{e}\Rightarrow R=e[/tex]
so the series converges absolutely for [tex]|x|\lt e[/tex] and diverges for [tex]|x|\gt e[/tex]. At the endpoints the root and ratio tests give the limit [tex]1[/tex], so they must be examined directly. For [tex]x=\pm e[/tex]
[tex]\left|a_nx^{n}\right|=\left(1-\frac{1}{n}\right)^{n^{2}}e^{n}=e^{\,n^{2}\ln\left(1-\frac{1}{n}\right)+n}[/tex]
By the Maclaurin expansion [tex]\ln(1-t)=-t-\frac{t^{2}}{2}-\frac{t^{3}}{3}-\ldots[/tex] with [tex]t=\frac{1}{n}[/tex], for [tex]n\ge 2[/tex]
[tex]n^{2}\ln\left(1-\frac{1}{n}\right)+n=-\frac{1}{2}-\sum_{k=3}^{\infty}\frac{1}{k\,n^{k-2}},\qquad 0\lt\sum_{k=3}^{\infty}\frac{1}{k\,n^{k-2}}\le\frac{1}{3}\sum_{j=1}^{\infty}\frac{1}{n^{j}}=\frac{1}{3(n-1)}\to 0[/tex]
Hence [tex]\left|a_nx^{n}\right|\to e^{-\frac{1}{2}}\ne 0[/tex], the necessary condition for convergence fails, and the series diverges at both [tex]x=e[/tex] and [tex]x=-e[/tex]. The interval of convergence is [tex]\left(-e, e\right)[/tex].

Solution:
For [tex]n\ge 2[/tex] we have [tex]n^p\gt 1[/tex], so all denominators are positive and the series is alternating. However, the sequence [tex]\frac{1}{n^p+(-1)^n}[/tex] is not monotone (its denominator is [tex]n^p+1[/tex] for even [tex]n[/tex] and [tex]n^p-1[/tex] for odd [tex]n[/tex]), so the Leibniz test cannot be applied directly. We split off the main term:
[tex]a_n=\frac{(-1)^n}{n^p+(-1)^n}=\frac{(-1)^n}{n^p}-c_n,\qquad c_n=\frac{1}{n^p\left(n^p+(-1)^n\right)}\gt 0[/tex]
The series [tex]\sum\frac{(-1)^n}{n^p}[/tex] converges for every [tex]p\gt 0[/tex] by the Leibniz test, because [tex]\frac{1}{n^p}[/tex] decreases to [tex]0[/tex]. For the positive series [tex]\sum c_n[/tex] we have
[tex]c_n\times n^{2p}=\frac{n^p}{n^p+(-1)^n}\to 1[/tex]
so by the limit comparison test [tex]\sum c_n[/tex] converges exactly when [tex]\sum\frac{1}{n^{2p}}[/tex] converges, i.e. when [tex]2p\gt 1[/tex].
If [tex]p\gt\frac{1}{2}[/tex], then [tex]\sum a_n[/tex] is the difference of two convergent series, hence it converges. If [tex]0\lt p\le\frac{1}{2}[/tex], the series diverges: otherwise [tex]\sum c_n=\sum\frac{(-1)^n}{n^p}-\sum a_n[/tex] would converge.
Hence the series converges exactly for [tex]p\gt\frac{1}{2}[/tex] (absolutely for [tex]p\gt 1[/tex] and conditionally for [tex]\frac{1}{2}\lt p\le 1[/tex]).

Solution:
Convergence: by the Cauchy root test, since [tex]\sqrt[k]{k}\to 1[/tex],
[tex]\sqrt[k]{\frac{k^{n}}{a^{k}}}=\frac{\left(\sqrt[k]{k}\right)^{n}}{a}\to\frac{1}{a}\lt 1[/tex]
so [tex]f(n)[/tex] is a convergent series of positive terms for every [tex]n[/tex]. We also put [tex]f(0)=\sum_{k=1}^{\infty}\frac{1}{a^{k}}=\frac{1}{a-1}[/tex] (a geometric series with ratio [tex]\frac{1}{a}\lt 1[/tex]). Multiply [tex]f(n)[/tex] by [tex]a[/tex] and shift the index, [tex]k=m+1[/tex]:
[tex]af(n)=\sum_{k=1}^{\infty}\frac{k^{n}}{a^{k-1}}=\sum_{m=0}^{\infty}\frac{(m+1)^{n}}{a^{m}}=1+\sum_{m=1}^{\infty}\frac{(m+1)^{n}}{a^{m}}[/tex]
By the binomial theorem [tex](m+1)^{n}=\sum_{j=0}^{n}\binom{n}{j}m^{j}[/tex]. Each of the finitely many series [tex]\sum_{m\ge 1}\frac{m^{j}}{a^{m}}=f(j)[/tex] converges, so we may sum term by term:
[tex]\sum_{m=1}^{\infty}\frac{(m+1)^{n}}{a^{m}}=\sum_{j=0}^{n}\binom{n}{j}f(j)=f(0)+\sum_{j=1}^{n-1}\binom{n}{j}f(j)+f(n)[/tex]
Hence, using [tex]1+\frac{1}{a-1}=\frac{a}{a-1}[/tex],
[tex]af(n)=\frac{a}{a-1}+\sum_{j=1}^{n-1}\binom{n}{j}f(j)+f(n)\ \Rightarrow\ (a-1)f(n)=\frac{a}{a-1}+\sum_{j=1}^{n-1}\binom{n}{j}f(j)[/tex]
which is the required equality. For [tex]n=1[/tex] it gives [tex]f(1)=\frac{a}{(a-1)^{2}}[/tex].

Solution:
Note that [tex]n^2+2n+4=(n+1)^2+3[/tex] and [tex]n^2+n+3=n(n+1)+\sqrt{3}\,\sqrt{3}[/tex]. Consider the plane vectors [tex]\vec{v}_n=\left(n, \sqrt{3}\right)[/tex]. Then [tex]\left|\vec{v}_n\right|=\sqrt{n^2+3}[/tex], [tex]\left|\vec{v}_{n+1}\right|=\sqrt{n^2+2n+4}[/tex], and the scalar product of [tex]\vec{v}_n[/tex] and [tex]\vec{v}_{n+1}[/tex] is [tex]n(n+1)+3[/tex]. So, if [tex]\varphi_n\in[0, \pi][/tex] is the angle between [tex]\vec{v}_n[/tex] and [tex]\vec{v}_{n+1}[/tex],
[tex]\cos\varphi_n=\frac{n(n+1)+3}{\left|\vec{v}_n\right|\,\left|\vec{v}_{n+1}\right|}=\frac{n^2+n+3}{\sqrt{n^2+3}\,\sqrt{n^2+2n+4}}[/tex]
and the [tex]n[/tex]-th term of the series is exactly [tex]\varphi_n[/tex]. Let [tex]\theta_n\in\left(0, \frac{\pi}{2}\right)[/tex] be the angle between [tex]\vec{v}_n[/tex] and the positive [tex]x[/tex]-axis; then [tex]\tan \theta_n=\frac{\sqrt{3}}{n}[/tex]. This decreases as [tex]n[/tex] grows, so [tex]\theta_n\gt\theta_{n+1}[/tex] and [tex]\varphi_n=\theta_n-\theta_{n+1}[/tex]. The partial sums telescope:
[tex]\sum_{n=1}^{N}\varphi_n=\sum_{n=1}^{N}\left(\theta_n-\theta_{n+1}\right)=\theta_1-\theta_{N+1}[/tex]
Here [tex]\tan \theta_1=\sqrt{3}[/tex], so [tex]\theta_1=\frac{\pi}{3}[/tex], and [tex]\tan \theta_{N+1}=\frac{\sqrt{3}}{N+1}\to 0[/tex], so [tex]\theta_{N+1}\to 0[/tex]. Hence the series converges to [tex]\frac{\pi}{3}[/tex] and the required value is
[tex]\frac{2016}{\pi}\times \frac{\pi}{3}=672[/tex]

Solution:
For [tex]0\lt x\le 1[/tex], by the Maclaurin series of [tex]e^u[/tex],
[tex]x^{-x}=e^{-x\ln x}=\sum_{k=0}^{\infty}\frac{\left(-x\ln x\right)^k}{k!}[/tex]
Let [tex]h(x)=-x\ln x[/tex] for [tex]0\lt x\le 1[/tex] and [tex]h(0)=0[/tex]; [tex]h[/tex] is continuous on [tex][0, 1][/tex], because [tex]x\ln x\to 0[/tex] as [tex]x\to 0^{+}[/tex]. From [tex]h'(x)=-\ln x-1[/tex] the maximum of [tex]h[/tex] is attained at [tex]x=\frac{1}{e}[/tex], so [tex]0\le h(x)\le\frac{1}{e}[/tex] on [tex][0, 1][/tex]. Hence
[tex]\left|\frac{h(x)^k}{k!}\right|\le\frac{1}{e^k\,k!},\qquad\sum_{k=0}^{\infty}\frac{1}{e^k\,k!}=e^{1/e}[/tex]
By the Weierstrass M-test the series of continuous functions [tex]\sum_{k=0}^{\infty}\frac{h(x)^k}{k!}[/tex] converges uniformly on [tex][0, 1][/tex]; its sum [tex]e^{h(x)}[/tex] equals [tex]x^{-x}[/tex] for [tex]x\gt 0[/tex] (and the limit value [tex]1[/tex] at [tex]x=0[/tex]). Therefore it may be integrated term by term:
[tex]\int_{0}^{1}x^{-x}\,dx=\sum_{k=0}^{\infty}\frac{1}{k!}J_k,\qquad J_k=\int_{0}^{1}\left(-x\ln x\right)^k dx[/tex]
In [tex]J_k[/tex] substitute [tex]x=e^{-t}[/tex], [tex]t\in\left(0, +\infty\right)[/tex]; then [tex]dx=-e^{-t}dt[/tex] and [tex]-x\ln x=t\,e^{-t}[/tex]. Then put [tex]u=(k+1)t[/tex]:
[tex]J_k=\int_{0}^{\infty}t^k e^{-kt}e^{-t}\,dt=\int_{0}^{\infty}t^k e^{-(k+1)t}\,dt=\frac{1}{(k+1)^{k+1}}\int_{0}^{\infty}u^k e^{-u}\,du=\frac{k!}{(k+1)^{k+1}}[/tex]
because [tex]\int_{0}^{\infty}u^k e^{-u}\,du=k![/tex] (integrate by parts [tex]k[/tex] times). Therefore
[tex]\int_{0}^{1}\frac{dx}{x^{x}}=\sum_{k=0}^{\infty}\frac{1}{k!}\times \frac{k!}{(k+1)^{k+1}}=\sum_{k=0}^{\infty}\frac{1}{(k+1)^{k+1}}=\sum_{n=1}^{\infty}\frac{1}{n^{n}}[/tex]
which was to be proved.

Solution:
The series [tex]S[/tex] has positive terms and is bounded above by [tex]\sum\frac{1}{n^{2}}[/tex], so it converges. Every natural number [tex]m[/tex] can be written in exactly one way as [tex]m=q\times k^{3}[/tex], where [tex]q[/tex] is cube-free and [tex]k\in\mathbb{N}[/tex]. Indeed, if [tex]m=\prod p^{e_p}[/tex] is the prime factorization, write [tex]e_p=3s_p+r_p[/tex] with [tex]r_p\in\{0, 1, 2\}[/tex] and take [tex]q=\prod p^{r_p}[/tex], [tex]k=\prod p^{s_p}[/tex]. Conversely, in any such representation the exponents in [tex]q[/tex] are [tex]0[/tex], [tex]1[/tex] or [tex]2[/tex], so [tex]r_p[/tex] and [tex]s_p[/tex] are the remainder and the quotient of [tex]e_p[/tex] divided by [tex]3[/tex], hence they are unique.
A series of positive terms may be rearranged and grouped arbitrarily, so (with [tex]q[/tex] running over the cube-free numbers) the double series factors as a product:
[tex]\frac{\pi^{2}}{6}=\sum_{m=1}^{\infty}\frac{1}{m^{2}}=\sum_{q}\sum_{k=1}^{\infty}\frac{1}{q^{2}k^{6}}=\left(\sum_{q}\frac{1}{q^{2}}\right)\left(\sum_{k=1}^{\infty}\frac{1}{k^{6}}\right)=S\times \frac{\pi^{6}}{945}[/tex]
Therefore
[tex]S=\frac{\pi^{2}}{6}\times \frac{945}{\pi^{6}}=\frac{315}{2\pi^{4}}\approx 1.617[/tex]

Solution:
Let [tex]G(x)=\sum_{n=0}^{\infty}F_n x^n[/tex]. By the recurrence,
[tex]G(x)-xG(x)-x^2G(x)=F_0+\left(F_1-F_0\right)x=x\ \Rightarrow\ G(x)=\frac{x}{1-x-x^2}[/tex]
By Binet's formula [tex]F_n=\frac{\varphi^n-\psi^n}{\sqrt{5}}[/tex], where [tex]\varphi=\frac{1+\sqrt{5}}{2}[/tex] and [tex]\psi=\frac{1-\sqrt{5}}{2}[/tex], the radius of convergence is [tex]\frac{1}{\varphi}=\frac{\sqrt{5}-1}{2}\approx 0.618\gt\frac{1}{2}[/tex], so at [tex]x=\frac{1}{2}[/tex] the series and its term-by-term derivatives converge. Put
[tex]H(x)=xG(x)=\sum_{n=0}^{\infty}F_n x^{n+1}=\frac{x^2}{1-x-x^2}[/tex]
Differentiating twice term by term, we get for [tex]x\ne 0[/tex]
[tex]H''(x)=\sum_{n=0}^{\infty}(n+1)n\,F_n x^{n-1}=\frac{2}{x}\sum_{n=0}^{\infty}T_nF_nx^{n}[/tex]
so the required sum is
[tex]\sum_{n=0}^{\infty}\frac{T_nF_n}{2^{n+2}}=\frac{1}{4}\sum_{n=0}^{\infty}T_nF_n\left(\frac{1}{2}\right)^n=\frac{1}{4}\times \frac{1}{4}H''\left(\frac{1}{2}\right)=\frac{1}{16}H''\left(\frac{1}{2}\right)[/tex]
A direct computation gives
[tex]H''(x)=\frac{2\left(1+3x^2-x^3\right)}{\left(1-x-x^2\right)^3},\qquad H''\left(\frac{1}{2}\right)=\frac{2\times \frac{13}{8}}{\left(\frac{1}{4}\right)^3}=\frac{13}{4}\times 64=208[/tex]
Hence the sum equals [tex]\frac{208}{16}=13[/tex].

Solution:
The integrand [tex]g(t)=\frac{1}{2}\ln\left|\cot \frac{t}{2}\right|[/tex] has only logarithmic singularities at [tex]t=0[/tex] and [tex]t=\pm\pi[/tex], which are integrable, so [tex]F[/tex] is defined and continuous on [tex][-\pi, \pi][/tex]. Since [tex]g[/tex] is even, [tex]F[/tex] is odd, so all [tex]a_n=0[/tex] (including [tex]a_0[/tex]).
Moreover [tex]F(\pi)=0[/tex]: the substitution [tex]t=\pi-s[/tex] turns [tex]\cot \frac{t}{2}[/tex] into [tex]\tan \frac{s}{2}=\frac{1}{\cot \frac{s}{2}}[/tex], so the integral [tex]\frac{1}{2}\int_{0}^{\pi}\ln\cot \frac{t}{2}\,dt[/tex] equals its own negative. Thus [tex]F(-\pi)=F(\pi)=0[/tex]. Integrating by parts (the boundary term vanishes because [tex]F(0)=F(\pi)=0[/tex]):
[tex]b_n=\frac{2}{\pi}\int_{0}^{\pi}F(x)\sin nx\,dx=\frac{2}{n\pi}\int_{0}^{\pi}F'(x)\cos nx\,dx=\frac{1}{n\pi}I_n,\qquad I_n=\int_{0}^{\pi}\ln\cot \frac{x}{2}\,\cos nx\,dx[/tex]
The substitution [tex]x=\pi-s[/tex] gives [tex]I_n=-(-1)^nI_n[/tex], so [tex]I_n=0[/tex] for even [tex]n[/tex]. For odd [tex]n[/tex] we integrate by parts once more, using [tex]\left(\ln\cot \frac{x}{2}\right)'=-\frac{1}{\sin x}[/tex] (the boundary terms vanish, since the logarithmic singularity is multiplied by [tex]\sin nx[/tex], which tends to [tex]0[/tex] linearly at [tex]0[/tex] and at [tex]\pi[/tex]):
[tex]I_n=\frac{1}{n}\int_{0}^{\pi}\frac{\sin nx}{\sin x}\,dx[/tex]
For [tex]n=2m+1[/tex] the identity [tex]2\sin x\cos 2jx=\sin (2j+1)x-\sin (2j-1)x[/tex] telescopes to
[tex]\frac{\sin (2m+1)x}{\sin x}=1+2\sum_{j=1}^{m}\cos 2jx[/tex]
whose integral over [tex][0, \pi][/tex] is [tex]\pi[/tex]. Hence [tex]I_n=\frac{\pi}{n}[/tex] and [tex]b_n=\frac{1}{n^2}[/tex] for odd [tex]n[/tex], while [tex]b_n=0[/tex] for even [tex]n[/tex]:
[tex]F(x)=\sum_{k=1}^{\infty}\frac{\sin \left((2k-1)x\right)}{(2k-1)^2}[/tex]
Since [tex]\sum\frac{1}{(2k-1)^2}[/tex] converges, the series converges uniformly (Weierstrass M-test) to a continuous function with the same Fourier coefficients as the continuous function [tex]F[/tex] (for which [tex]F(-\pi)=F(\pi)[/tex]); two such functions coincide, so the equality holds on the whole interval [tex][-\pi, \pi][/tex].