How to prove that    $$\sum_{k\geq 1} \frac{1}{2^k k^2}=\frac{\pi^2}{12}-\frac{1}{2}\log(2)^2$$  without using the well-known $\operatorname{Li}_2\left( \frac{1}{2} \right)$ ?  Edited : Thanks for L.F , but I should have made it clear that I want an answer using only series manipulations .
123540
My teacher posed an infinite series question to me today and I'm not quite sure how to start to go about it. $$\sum_{r=1}^{\infty} \dfrac{2^{-r}}{r^{2}}$$ Any hints would be much appreciated.
35641
I have the following problem:  Prove that if $k &gt; 1$, then $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{4k}{(k^2 - 1)^2}$$ Hence simplify $$\sum\limits_{k = 2}^n \dfrac{k}{(k^2 - 1)^2}$$  My Work: $$\dfrac{1}{(k - 1)^2} - \dfrac{1}{(k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k - 1)^2 (k + 1)^2} = \dfrac{(k + 1)^2 - (k - 1)^2}{(k^2 - 2k + 1)(k^2 + 2k + 1)}$$ I could have continued and multiplied the denominator out, but, at this point, I'm thinking that I'm actually expected to have proceeded differently, rather than multiplying the denominator out fully? And I don't have solutions for this problem, so I can't check anything. I would appreciate it if someone could please take the time to explain this problem.
115573
Let $n \in \mathbb{Z}^+$, prove the identity $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$$  First of all  $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)$$ $$=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)$$ $$=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k}  \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)$$  We have for the first sum $$(1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.$$ For the second sum  $$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$  Integrating both sides from $0$ to $x$, we see that   $$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.$$  Putting $x=1$ yields $$\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}$$  Here where I have stopped. I could not get them similar for what I have. Would someone help me out !
30738
How tipe are this series?  $$\sum_{n=2}^\infty \frac 1 {n^2\ln n}$$ But $$\sum_{n=2}^\infty \frac 1 {n\ln n\ln(\ln n)}$$ I used Condensation Test, but I stack after that. Thanks!
103332
Can anyone help me with this?  What is the sum of this series:  $\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}$  I got it after plugging $x=-1$ in a Fourier series   Thank you!
80310
How can I prove $$\sum_{i=1}^{n}(-1)^{i-1}\dbinom{n}{i}\frac{1}{i}=\frac{1}{1}+\frac{1}{2}+....+\frac{1}{n}\quad ?$$ I have tried induction, but it doesn't work.
22882
How to prove that  $$\sum_{n=1}^{\infty}\frac{1}{4^n\cos^2\frac{x}{2^n}}=\frac1{\sin^2x}-\frac1{x^2}$$
8724
Let $S(n)$ be $S(n)=\left\{k\;\left|\;\left\{\frac{n}{k}\right\}\right.\geq \frac{1}{2}\right\}$,where $\{x\}$ is the fractional part of $x$  Prove that :  \begin{align} \sum_{k\in S(n)} \varphi(k)=n^2 \end{align}  maybe the fact \begin{align} \sum_{k \leq 2n}\left\lfloor{\frac{n}{k}+\frac{1}{2}}\right\rfloor \varphi(k)=\frac{3n^2+n}{2}\end{align}  and \begin{align} \sum_{k \leq 2n}\left\lfloor{\frac{n}{k}}\right\rfloor \varphi(k)=\frac{n^2+n}{2}\end{align}  can help, but I can't prove these two equations    Edited:Sorry that my question is to prove the two equation before, because I've already known how to relate the problem with these two equations, and I've also known how to prove the second equation from  but I can't solve the first one with the similar method.
89519
So I have been trying for a few days to figure out the sum of   $$ S = \sum_{k=1}^\infty \frac{1}{k^2 - a^2} $$ where $a \in (0,1)$. So far from my nummerical analysis and CAS that this sum equals  $$ S = \frac{1}{2a} \left[ \frac{1}{a} \, - \, \pi \cot(a\pi) \right] $$  But I have not been able to prove this yet. Anyone know how? My guess is that the  sum of this series is related to fourier-series but nothing particalr comes to mind.   For the easy values, I have been able to use telescopic series, and a bit of algebraic magic, but for the general case I am stumpled. Anyone have any ideas or hints? Cheers.
27799
Can someone tell me how to get a closed form for  $$\sum_{k=1}^n k^p$$  For $p = 1$, it's just the classic $\frac{n(n+1)}2$.   What is it for $p &gt; 1$?
33465
I've been trying to prove, by arithmetical means, that  $$\sum_{k=1}^\infty \frac{((k-1)!)^2}{(2k)!} =\frac{1}{3}\sum_{k=1}^{\infty}\frac{1}{k^{2}}$$  without success.  When I say "by arithmetical means" I mean to say, go from the left to the right expression just by symbolic manipulation.  Can anyone devise a way of doing this?
4231
I am being asked to prove that $$\sum\limits_{k=0}^{n}\cos(kx)=\frac{1}{2}+\frac{\sin(\frac{2n+1}{2}x)}{2\sin(x/2)}$$ I have some progress made, but I am stuck and could use some help. What I did: It holds that $$\sum\limits_{k=0}^{n}\cos(kx)=\sum\limits_{k=0}^{n}Re(\cos(kx))=\sum\limits_{k=0}^{n}Re(\cos(x)^{k})=Re(\sum\limits_{k=0}^{n}\cos(x)^{k})=Re\left(\cos(0)\cdot\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right) $$ For any $z_{1},z_{2}\in\mathbb{C}$ we have it that if $z_{1}=a+bi,z_{2}=c+di$ then $$\frac{z_{1}}{z_{2}}=\frac{z_{1}\overline{z2}}{|z_{2}|^{2}}=\frac{(a+bi)(c-di)}{|z_{2}|^{2}}=\frac{ac-bd+i(bc-ad)}{|z_{2}|^{2}}$$ hence $$Re\left(\frac{z_{1}}{z_{2}}\right)=\frac{Re(z_{1})Re(z_{2})-Im(z_{1})Im(z_{2})}{|z_{2}|^{2}}$$ Thus, $$Re\left(\frac{\cos(x)^{n}-1}{\cos(x)-1}\right)=\frac{(\cos(nx)-1)(\cos(x)-1)-\sin(nx)\sin(x)}{(\cos(x)-1)^{2}+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{\cos^{2}(x)-2\cos(x)+1+\sin^{2}(x)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2\cos(x)+2}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(\cos(x)-1)}= \frac{=\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{-2(-2\cdot\sin^{2}(x/2))}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(nx)\cos(x)-\cos(nx)-\cos(x)+1-\sin(nx)\sin(x)}{4\sin^{2}(x/2)}=\frac{\cos(x(n+1))-\cos(nx)-\cos(x)+1}{4\sin^{2}(x/2)} $$ This is the part where I am stuck, I would appriciate any help or hint on how to continue. Edit: Given the corrections by André I get: $$(\cos(nx+x)-1)(\cos(x)-1)+\sin(nx+x)\sin(x)=\cos(nx+x)\cos(x)-\cos(nx)-\cos(x)+1+\sin(nx+x)\sin(x)$$ so $$\cos(nx+x)\cos(x)+\sin(nx+x)\sin(x)=\cos(xn+x-x)-\cos(nx)=0$$ Edit 2: I found anoter mistake in the above, I will try to correct Edit 3: When multiplying correctly the above it works out :-)
54750
Show that $$\sum\limits_{k=1}^n \frac{1}{k} \geq \log(n)$$ Use this to deduce that the series $$\sum\limits_{k=1}^\infty \frac{1}{k}$$ diverges.   Hint: Use the estimate $$\frac{1}{k} \geq \int_k^{k+1} \frac{1}{x}\,\mathrm{d}x$$   Note that all instances of "$\log$" mean the natural logarithm, the inverse of the exponential. After doing the integration, I've concluded that $\frac{1}{k} \geq \log(1+\frac{1}{k})&gt;-\log(k)$  I suspect I'm supposed to use the hint to arrive at the first inequality, and then since $\log(n)$ can become arbitrarily large as $n$ increases, so must the harmonic series (essentially the comparison test).  I'm having trouble making this jump. Hints are appreciated, but please no solutions.
35282
I am trying to find a general summation for $$\sum_{k=1}^{n} \frac{(k-1)2^k}{k(k+1)}$$ I tried to expand it and look for some pattern. $$\frac{(1-1)2^1}{1(1+1)} + \frac{(2-1)2^2}{2(2+1)}+...+\frac{((k-1)-1)2^{k-1}}{(k-1)((k-1)+1} + \frac{(k-1)2^k}{k(k+1)}$$ If it can be solved through telescoping, I couldn't see how.
40528
$\sum_{i=1}^n(a_i)\sum_{i=1}^n(\frac{1}{a_i})\ge n^2$ is what I want to prove, where $a_i&gt;0$ and I need to showwhen equality holds.  I have tried using Cauchy-Schwarz, where inside the sums I multiply each value by 1 to do it but it doesn't work out.  I have also done  $\sum_{i=1}^n(a_i)\sum_{i=1}^n(\frac{1}{a_i})=\sum_{i=1}^n \sum_{i=1}^n(\frac{a_i}{a_i})=\sum_{i=1}^n\sum_{i=1}^n(1)=\sum_{i=1}^n(n)=n^2$, but this proves an equality.  Am I not allowed to do the first step I did in my work?
111461
I need to use the identity $\frac{1}{k^2-1}=\frac{1}{2}(\frac{1}{k-1}-\frac{1}{k+1})$ to evaluate $\sum_{k=2}^n \frac{1}{k^2-1}$.  I am confused about how to begin this proof.  Also, how am I supposed to use the identity here?  Do I need to use induction?
112289
If $x \geq 2$, then how do we prove that $$\int_{2}^{x} \frac{\mathrm dt}{\log^{n}{t}} = O\Bigl(\frac{x}{\log^{n}{x}}\Bigr)?$$