How to find the sum the the following series $\displaystyle\sum_{n=1}^{\infty}\sum_{k=1}^{\infty}\frac{1}{n^2k^2(n+k)^2}$ .I am absolutely clueless . I have tried to represent it as an integral but failed. Answer seems to be $\frac{\pi^6}{2835}$.
121100
While Solving some integral problem, I encountered the following infinite series:  $$\displaystyle \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n^2k^2(n+k)^2}$$  I have tried many methods including partial fractions... I seek help! Please provide hints if you don't have the complete answer.
60070
How can we evaluate this series  $$\sum_{n=1}^\infty \frac{n^a}{b^n}?$$  Here $a$ and $b$ are positive integers. If $b=1$ then series will be diverging, in other cases, it will be converging, but how to find this sum?  Isn't there a simple solution which doesn't involve Stirling numbers?
28868
What would be the sum of the following series?  $$\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^n}$$  Thanks
23497
How do I evaluate this?  $$\lim_{n\to \infty }\sum_{k=1}^{n}\frac{k}{k^2+n^2}$$  I got concerned for that, I've tried make it integral for Riemann but it still undone.
69564
How can we find the exact value of the infinite sum  $$ \displaystyle\sum_{n=1}^\infty \left\{\mathrm{e}-\Big(1+\frac1n\Big)^n\right\}? $$  This problem appears in:  , p.114.
6112
Find the sum $ \displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2}(2n-1)}$  $= 1 + \dfrac{1}{3\cdot 2^{2}} + \dfrac{1}{5\cdot 3^{2}} + \dfrac{1}{7\cdot 4^{2}} \ldots$  Then, I'm stuck. I didn't find any pattern to solve.  I just need a hint.
28799
$$\sum_{k=1}^{\infty} \frac{1}{5k+2}$$   Can I say that $$\sum_{k=1}^{\infty} \frac{1}{5k+2}=\sum_{k=1}\frac{1}{5k}+\sum_{k=1}^{\infty} \frac{1}{2}=\frac{1}{5}\sum_{k=1}^{\infty}\frac{1}{k}+\frac{1}{2}$$  and we know that $\\sum_{k=1}^{\infty} \frac{1}{k}$ diverges so the expression diverges too?  Or should I use the integral test $$\int_{1}^{\infty} \frac{1}{5k+2}=\frac{ln|5k+2|}{5}$$ from 1 to $\infty$=  $\infty$
59019
I am very confused about how to compute  $$\sum_{n=0}^{\infty}nk^n.$$  Can anybody help me?
34849
I'm trying to show that the series $\sum_{n=1}^\infty\frac{n}{n^2+1}$ is divergent.   I'm trying to use a comparison test to do so. To do so, I first can see that $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{n=1}^\infty\frac{n}{n^2+n} = \sum_{n=1}^\infty\frac{1}{n+1}$ And I can use a change of variable $k = n+1$, so show $\sum_{n=1}^\infty\frac{1}{n+1} = \sum_{k=2}^\infty\frac{1}{k} = \infty$. Then since $\sum_{n=1}^\infty\frac{n}{n^2+1} \geq \sum_{k=2}^\infty\frac{1}{k}$ it must be that $\sum_{k=2}^\infty\frac{1}{k}$ also diverges to infinity. Is this a valid way to show it?
91606
I'm supposed to determine whether this sum diverges or converges and if it converges then find its value:  $$ \sum_{n=2}^\infty \frac{1}{n^2-1}. $$  Using the comparison test I eventually showed that this converges.  But I can't figure out how to show what this sum converges to.  The only sums we actually found values for in my notes are geometric series which this clearly isn't.  I saw that I could use partial fraction decomposition to represent the terms as  $$\frac{1/2}{n-1}- \frac{1/2}{n+1} $$ but that just gets me $\infty - \infty$, so this isn't the way to do it.  I'm not sure how to find the value of this sum.  I don't need the full solution but a hint would be appreciated. Thanks. :)
86635
I try to solve the following sum:  $$\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{k^2n+2nk+n^2k}$$  I'm very curious about the possible approaching ways that lead us to solve it. I'm not experienced with these sums, and any hint, suggestion is very welcome. Thanks.
46049
I want to Calculate $\sum_{k=1}^n k2^{n-k}$. Here's my attempt:  $$\begin{align}  &amp;\sum_{k=1}^n k2^{n-k} = \sum_{k=1}^n k2^{n+1}2^{-k-1} =2^{n+1}\sum_{k=1}^n k2^{-k-1} &amp; &amp;\text{(1)} \\ &amp;k2^{-k-1} = -\frac{d}{dk}2^{-k} &amp; &amp;\text{(2)} \end{align}$$   Plugging $(2)$ to $(1)$,   $$\begin{aligned}  2^{n+1}\sum_{k=1}^n k2^{-k-1}  &amp;= 2^{n+1}\sum_{k=1}^n-\frac{d}{dk}2^{-k} \\  &amp;= -2^{n+1}\sum_{k=1}^n\frac{d}{dk}2^{-k} \\  &amp;= -2^{n+1}\frac{d}{dk}\sum_{k=1}^n2^{-k} \\  &amp;= -2^{n+1}\frac{d}{dk}(1-2^{-n}) \\ &amp;= -2^{n+1}\cdot 0 \\ &amp;= 0 \end{aligned}$$   Obviously, I don't know much about summation and derivative. Can you please let me know where I am doing wrong and give me a hint for this question? Thanks!
106805
So I just started learning series, and I am having trouble understanding them. I have a series below:  $$\sum_{n=0}^{\infty} \frac{1}{(2n+1)2^n}$$  Can someone explain to me how I should go about this, from start to finish? I couldn't find any online websites that explained how to find sums, so if you have that, it would be greatly appreciated!
29310
Find the sum to $n$ terms as well as the sum to infinity of the series:  $$\frac{1}{3} + \frac{1}{4\cdot 2!} + \frac{1}{5\cdot 3!} +\cdots.$$  I was trying this question many times but I didn't get the answer. I was using geometric series formula, but it was not applicable here...I don't know where I have to start, and I don't know any hint about this series....  If anybody helps me, I would be very thankful to him.
45468
How to compute the sum of the following series. I have no idea...  The series is $$\sum_{n=1}^\infty(1+1/2+\cdots+1/n)x^n.$$
64621
Can anyone tell me about the sum of the series $$\sum_{n=1}^\infty \frac{1}{(2n)(2n+1)}?$$ This is not a usual telescoping sum in which all the terms cancel out.
15471
Show that $\sum_{n = 1}^{+\infty} \frac{n}{2^n} = 2$. I have no idea to solve this problem. Anyone could help me?