Calculate sum $$\sum_{n=1}^\infty {n\over3^{n-1}}$$The result is $9\over 4$ but I don't know how to get that.
32926
I am trying to find a closed form solution for $\sum_{n\ge0} nx^n\text{, where }\lvert x \rvert&lt;1$.  This solution makes sense to me:  $\sum_{n\ge0} x^n=(1-x)^{-1} \\ \frac{d}{d x} \sum_{n\ge0} x^n = \frac{d}{d x} (1-x)^{-1} \\ \sum_{n\ge0} nx^{n-1} = (1-x)^{-2} \\ x \sum_{n\ge0} n x^{n-1} = x(1-x)^{-2} \\ \sum_{n\ge0} nx^n=\frac x{(1-x)^2}$  However, a  I am reading used the following method:  $$\sum_{n\ge0}nx^n=\sum_{n\ge0}x\frac d{dx}x^n= x\frac d{dx}\sum\limits_{n\ge0}x^n=x\frac d{dx}\frac1{1-x}=\frac x{(1-x)^2}$$  This seems closely related to the solution I described above, but I am having difficulty understanding it.  Can someone explain the method being used here?
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.
100546
Everyone knows about the classic  $$ \sum_{i=1}^{\infty} \dfrac{1}{2^i} = 1 $$ However, is there any way to find $$ \sum_{i=0}^{\infty} \dfrac{1}{2^{2^i}} = \dfrac12 + \dfrac14 + \dfrac{1}{16} + \dfrac{1}{256} + \cdots $$
52305
I was doing some problems in algebraic number theory and this series came up  $$\sum_{n=0}^\infty\frac{1}{9n^2+9n+2}.$$  So, I would like to know the value of this series. However, I don't want a full answer, only hints. I have tried to use Fourier's series, but with no success, perhaps an answer can be gotten usign this.  Thanks
71396
I have the following problem: $$\sum^{\infty}_{n=1}\dfrac{2}{\left(n+2\right)\sqrt{n}+n\sqrt{n+2}}$$ I should find the sum of this sequence. I tried to simplify but it does not work.
88825
$\sum_{n=1}^\infty n*(n+1)*x^n$  Hello everyone,  I need help in solving the question above.  I started with $\frac{1}{1-x} = \sum_{n=0}^\infty x^n$, if differentiated it once so it became $\frac{1}{(1-x)^2} =\sum_{n=1}^\infty nx^{n-1} $ but from here I don't know how to continue. Thanks.
100399
If the $n^{th}$ partial sum of a series $\sum a_n$ is $S_n = \dfrac{n-1}{n+1}$, find $a_n$ and the sum.  By definition,  the sum of the series is the $\lim n\to\infty$ of it's $n^{th}$ partial sum.  $$\text{ Sum = } \lim_{n\to\infty} = \dfrac{n-1}{n+1} = 1$$  I am asked to find $a_n$. How do I do this? What is the procedure?
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
5245
I have to find the sum of the following series:  $$\sum\limits_{n=1}^\infty \frac{n^3}{n!}$$  I know how to prove the convergence of this series, but how do i find the sum. I can't use the properties of derivation or integration of the series, i have to do it using the definition that a series converges iff it's sequence of partial sum converges.
101345
I tried getting it into a closed form but failed. Could someone help me out?  $$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$
59019
I am very confused about how to compute  $$\sum_{n=0}^{\infty}nk^n.$$  Can anybody help me?
19671
Calculate $$\sum_{k=1}^{\infty}\frac{1}{2^{k+1}-1}.$$    I used Wolfram|Alpha to compute it and got it to be approximately equal to $0.6$. How to compute it? Can someone give me a hint or a suggestion to do it?
16813
I would like to know: How come that  $$\sum_{n=1}^\infty n x^n=\frac{x}{(x-1)^2}$$  Why isn't it infinity?
118828
$$\sum_{n=1}^\infty \frac{1}{n^2 2^n}$$  I am new in series thus I tried a pair of methods to compute but I couldn't
121291
I have a clueless friend who believes that   $$ \sum_{n=1}^\infty \frac{1}{2^n} $$  doesn't equal $1$ in the 'normal arithmetical sense'.   He doesn't believe that this series flat out equals $$1$$ Is he correct?
113978
I am having trouble seeing how $\sum_{n=1}^{\infty}\frac{2}{(2n+1)(2n+3)}$ equals $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. I can see $\sum_{n=1}^{\infty}\frac{1}{2n+1}+\frac{1}{2n+3}$ but not $\sum_{n=1}^{\infty}\frac{1}{2n+1}-\frac{1}{2n+3}$. All help is appreciated, thanks.
1602
How can I evaluate $$\sum_{n=1}^\infty\frac{2n}{3^{n+1}}$$? I know the answer thanks to , but I'm more concerned with how I can derive that answer. It cites tests to prove that it is convergent, but my class has never learned these before. So I feel that there must be a simpler method.  In general, how can I evaluate $$\sum_{n=0}^\infty (n+1)x^n?$$