I want to show (or rather, know why) $$\sum_{k = 1}^{\infty} \frac{k^2}{k!} = 2e.$$ I have tried using the power series expansion for $e^x$, and also series manipulations with $\ln(x)$, but with no success. Any hints would be appreciated.
52085
How can I find the value of the sum $\sum_{k=1}^{\infty}\frac{k^n}{k!}$? for example for $n=6$, we have $$\sum_{k=1}^{\infty}\frac{k^6}{k!}=203e.$$
100131
I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$  $$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$  I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$  so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$  But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?  Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
24341
Evaluate $\sum_{k=1}^{\infty} \frac{1}{k^2(k+1)^2}$.   I've tried to use a telescoping approach, but it doesn't seem to work. It doesn't also seem to converge to a "pretty" value as well.
97761
What is the function represented by the power series $$ \sum_{k=2}^\infty\frac{x^k}{k(k-1)}\quad? $$  It looks like $\dfrac1{1-x}$ but I don't know.
101295
I need to study this sum: $\sum_{n=1}^{\infty} \frac{n!\, i^n}{n^n}$. Taking: $$\lim_{n\rightarrow \infty} \; \left|\left(\frac{n!\: i^n}{n^n}\right)^{\frac{1}{2}}\right|$$ $$\rightarrow \lim_{n\rightarrow \infty}  \left|\left(\frac{n!}{n^n}\right)^{\frac{1}{2}}\right|$$  Using Stirling approximation considering the limit:  $$\ln n! \approx n \ln\:n-n$$ then $n! \approx (\frac{n}{e})^n$ (is this correct?):  $$\rightarrow \lim_{n\rightarrow \infty} \left|\frac{n}{n\, e}\right| = \frac{1}{e}$$ This doesn't make too much sense because I know the series diverges.  I think I'm missing a $\sqrt{2\pi n}$ in Stirling approximation but I don't understand why that pops out taking the $e^{(\;)}$ from the first expression.
57644
Finding sum of series $\displaystyle \sum^{\infty}_{k=0}\frac{(k+1)(k+2)}{2^k}$   what i try  Let $\displaystyle x=\frac{1}{2},$ Then series sum is $\displaystyle \sum^{\infty}_{k=0}(k+1)(k+2)x^k$  $\displaystyle \sum^{\infty}_{k=0}x^k=\frac{1}{1-x}\Rightarrow \sum^{\infty}_{k=0}x^{k+1}=\frac{x}{1-x}=\frac{1-(1-x)}{1-x}$  differentiating both side  $\displaystyle \sum^{\infty}_{k=0}(k+1)x^{k}=+\frac{1}{(1-x)^2}$  How do i solve it help me please
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!
91354
Show that $\sum_{k=1}^\infty \frac{\ln{k}}{k^2}$ converges or diverges   I know that I must use the Limit Comparison test and my instinct tells me that this series will converge. I cannot, however, think of another series $b_k$, to compare my given series to in order to show that it converges.  Any hints?   Thank you!
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?
113204
How could the series below be computed ? $$\sum_{k=0}^{\infty}\frac{1}{2^{k!}}$$ It's not a series from a book, but a series I thought of many times, and I didn't manage to figure out what I should do here. I'm just curious to know if there are some known ways for approaching such a series. Thanks!
79373
Could anyone help me find an explicit formula for:  $$ \sum_{n=1}^\infty n^2x^n $$  We're supposed to use: $$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}  \qquad  |x| &lt;1 $$
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?
13596
I can not find the function from which I have to start to calculate this power series.  $$\sum_{n=1}^\infty nx^n$$ Any tips?. Thanks.
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?
75893
This is the sum   $$\sum_{n=1}^{\infty}\frac{1}{{5^n}^!}$$  My first attempt was to assume that the series does converge to a rational number $a/b$. But the $n!$ bothered me and I failed in my proof.  How would you try to prove this series?
9386
Can you please explain why $$ \sum_{k=1}^{\infty} \dfrac{k}{2^k} = \dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots = 2 $$  I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
38325
Show that this series $$\sum_{n=1}^{\infty}\dfrac{e^n\cdot n!}{n^n}$$ is divergent. My try: since $$u_{n}=\dfrac{e^n\cdot n!}{n^n},\Longrightarrow \dfrac{u_{n+1}}{u_{n}}=\dfrac{e^{n+1}\cdot(n+1)!}{(n+1)^{n+1}}\cdot\dfrac{n^n}{e^n\cdot n!}=\dfrac{e}{\left(1+\dfrac{1}{n}\right)^n}$$ then $$\lim_{n\to\infty}\dfrac{u_{n+1}}{u_{n}}=\lim_{n\to\infty}\dfrac{e}{\left(1+\dfrac{1}{n}\right)^n}=1$$ so this limit is $1$, therefore I can't prove it by using the Ratio test.   From sos440 suggestion: by using Stirling's approximation, we have $$\dfrac{e^n\cdot n!}{n^n}\approx \sqrt{2n\pi}\to\infty.$$  Maybe this problem requires other methods.