I want to use $S = \sum_{k=0} ^\infty (-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ and get the relation $(a^2+1) S'=1−aS$. So far I am just getting $\frac{dS}{da} = \sum_{k=0} ^\infty (-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k} (2k+1)$, which I am not seeing how to use with S to get that relation. What should I do to progress and get the desired relation?
124130
I want to use $S = \sum_{k=0} ^\infty (-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k+1}$ and get the relation $(a^2+1) S'=1−aS$. So far I am just getting $\frac{dS}{da} = \sum_{k=0} ^\infty (-1)^k \frac{(2k)!!}{(2k+1)!!} a^{2k} (2k+1)$, which I am not seeing how to use with S to get that relation. What should I do to progress and get the desired relation?
91515
$a_n=(-1)^{n-1}, \;  s_n=\sum_{i=1}^{n}a_i$ then find  $ \lim_{n\to &gt; \infty}\frac{s_1+s_2+\dots s_n}{n}$   $$s_k=1,\; \text{if k is odd and } s_k=0  \text{ if k is even}  $$  Cauchy's theorem for a sequence $(x_n) $ in $R$, we have $\lim\frac{x_1+x_2+\dots x_n}{n}=\lim x_n$  How do I make use of this theorem here when $(s_n)$ is oscillating between $0$ and $1$?
101687
Find the value of $$S=\sum_{n=1}^{\infty}\left(\frac{2}{n}-\frac{4}{2n+1}\right)$$    My Try:we have  $$S=2\sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{2}{2n+1}\right)$$    $$S=2\left(1-\frac{2}{3}+\frac{1}{2}-\frac{2}{5}+\frac{1}{3}-\frac{2}{7}+\cdots\right)$$ so  $$S=2\left(1+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\cdots\right)$$  But we know  $$\ln2=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$  So  $$S=2(2-\ln 2)$$  Is this correct?
8533
Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$    I think this is known (see ), I appreciate any hint or link for the solution (or the full solution).
33804
$$\sum_{r=1}^nr(r+1)=\frac{n(n+1)(n+2)}{3}$$  could you help me with how exactly I work this out?
69259
If $\displaystyle S=\sum^{n}_{k=0}(-1)^k\frac{1}{k+m+1}\binom{n}{k}$ and f $\displaystyle T=\sum^{m}_{k=0}(-1)^k\frac{1}{k+n+1}\binom{m}{k}$.Then $S-T$ is   Try:$$S=\sum^{n}_{k=0}(-1)^k\int^{1}_{0}x^{k+n}\binom{n}{k}dx=\int^{1}_{0}x^n\sum^{n}_{k=0}(-x)^k\binom{n}{k}dx$$  Could some help me, thanks
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 $\infty$=  $\infty$
15185
This problem comes from another equation on another question (. I tried to split it in half but I found out that $$\sum_{k=0}^\infty \frac{k}{2^k}$$ can't be divided. Knowing that $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$ I wrote that $$\sum_{k=0}^\infty \frac{k}{2^k}=\sum_{k=0}^\infty \left(\frac{\sqrt[k] k}{2}\right)^k=\frac{1}{1-\frac{\sqrt k}{2}}=\frac{2}{2-\sqrt[k] k}$$ But that's not what I wanted. Could anyone help me?
107530
Find sum of the series $$\sum_{n=1}^{\infty}\frac{x^{3n}}{\left(3n\right)!}$$ using differentiation. So far I found that $$S(x)+1=S'''(x)$$ but it does not help. Then I tried different interesting ideas  like $$S(x)+S'(x)+S''(x)=e^x-1\,.$$Maybe if I get the third equation it will allow me to construct a kind of differential equation. Then, by solving it, obtain $S(x)$.
17075
If $$S=\frac{1}{1+1^2+1^4}+\frac{2}{1+2^2+2^4}+\cdots+\frac{n}{1+n^2+n^4}\,$$ find the value of $14S$.  The question can be simplified to:  Find $S=\sum\limits_{k=1}^n\,t_k$ if $t_n=\dfrac{n}{1+n^2+n^4}$.  As $1+n^2+n^4$ forms a GP, $$t_n=\frac{n(n^2-1)}{n^6-1}\,.$$ But I can't figure out how to solve further. It would be great if someone could help.
104205
I'm trying to find the sum of :   $$\sum_{i=1}^n\frac i {2^i}$$  I've tried to run $i$ from $1$ to $∞$ , and found that the sum is $2$ , i.e :   $$\sum_{i=1}^\infty\frac i {2^i} =2$$  since :  $$(1/2 + 1/4 + 1/8 + \cdots) + (1/4 + 1/8 + 1/16 +\cdots) + (1/8 + 1/16 + \cdots) +\cdots = 1+ 1/2 + 1/4 + 1/8 + \cdots = 2 $$  But when I run $i$ to $n$ , it's a little bit different , can anyone please explain ?   Regards
52877
So I've been struggling with this sum for some time and I just can't figure it out. I tried proving by induction that if the sum above is a $S_n$ then $S_{n+1} = 4S_n$, but I didn't really succeed so here I am. Thanks in advance.
107744
Here is the series:  $$ \sum_{k = 0}^{\infty} \frac{(-1)^k}{2k + 1}. $$  I don't know how to start at all. Thank you for your help.
2819
$$\sum_{k=2}^{\infty}\frac{1}{k^2-1}$$   I found that:  $$\sum_{k=2}^{\infty}\frac{1}{k^2-1}=\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$$  and $\sum_{k=2}^{\infty}\frac{1}{2(k-1)}-\frac{1}{2(k+1)}$ is a telescopic series so we need $lim_{n \to \infty} \frac{1}{2}-\frac{1}{2(n+1)}=\frac{1}{2}$ but the answer is $\frac{3}{4}$
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 $p &gt; 1$?
79495
I am looking at some homework where I have:  $\sum_{n=1}^\infty \frac{1}{n^2(n+1)}$  How can I sum this?   I know that   $\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$  and also that  $\sum_{n=1}^\infty \frac{1}{n(n+1)}=1$  But I just can't see or find the connection
57076
Does anyone have any idea how to prove that  $$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)...(n+k)} = \frac{1}{kk!}$$