I know that the Series $\sum_{n=1}^\infty n \cdot x^n $ converges to $\frac{x}{(x-1)^2}$ but I'm not sure how to show it. I'm pretty sure that has been asked before, but I wasn't able to find anything...
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 $$\sum_{n=0}^\infty (n+1)x^n?$$
75491
Convergent, absolutely convergent or divergent:   $$\sum_{k=1}^{\infty}\frac{2^{k}\cdot k!}{k^{k}}$$   I have used ratio test because we got a fraction here and I think I did alright till the end:  $$\lim_{k\rightarrow \infty}\frac{\frac{2^{k+1}\cdot (k+1)!}{(k+1)^{k+1}}}{\frac{2^{k}\cdot k!}{k^{k}}} = \lim_{k\rightarrow \infty}\frac{2^{k+1}\cdot (k+1)! \cdot k^{k}}{(k+1)^{k+1}\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2^{k}\cdot 2^1\cdot k! \cdot (k+1)\cdot k^{k}}{(k+1)^{k}\cdot (k+1)\cdot 2^{k}\cdot k!} = \lim_{k\rightarrow \infty}\frac{2k^{k}}{(k+1)^{k}}$$ $$=2\lim_{k\rightarrow \infty} \left( \frac{k}{k+1}\right)^{k}$$  Now I don't know (without a calculator) to what this would converge / diverge to... In the exam we are not allowed to use a calculator... So what to do?  The denominator will be greater than the enumerator by 1, so dividing each other we got something $&lt;1$. We take exponent $k$ which is $\geq 1$, so we will end up with $&lt;1$ again. Multiply this with 2 we get something $&lt; 1$ but $&gt; 0$ and thus the series is convergent...?  I hope I have described it well? Would you give me full points to this task? :D  Edit: I haven't described it well in the end. See the accepted answer and its comments! Thanks a lot to everyone - from every question I ask here, I always learn new things :-)
99381
I'm trying to prove this sequence converges:  $\sum_{n=1}^{\infty}  \frac{1}{n\ln^2(n)}$  I noticed that this is continuous function which its derivative is always less than $0$ for $ x \gt 1 $, so I tried to do the integral test but with no success...   Some help?  Thanks!
113494
Does the series $\sum_{n=1}^{\infty}((1/n)-\sin(1/n))$ converge..? Can anyone please give me a simple proof..
69751
Find the radius of convergence of $\sum_{n=1}^{\infty}{n!x^{n!}}$. Should I look at this series as: $\sum_{n=1}^{\infty}({n!x^{(n-1)!})x^{n}}$? I am really confues here. In addition, any attempt to compute any kind of "lim sup" fails. I would appreciate your help in this.
109401
This is not a duplicate of this .  The linked question says that it suffices to show that if $(x_n)\to x$ then $(\frac{x_1+\cdots+x_n}{n})\to x$ to prove my question, but how so? I tried using the same strategy as how one proves that if $(x_n)\to x$ then $(\frac{x_1+\cdots+x_n}{n})\to x$, by "splitting" the product in th $N$th term:  $$\sqrt[n]{x_1x_2\cdots x_n}=\sqrt[n]{x_1x_2\cdots x_Nx_{N+1}\cdots x_n}=\sqrt[n]{x_1x_2\cdots x_N} \sqrt[n]{x_{N+1}\cdots x_n}$$  but it seems I can't use for now the definition of convergence of $(x_n)$ because of the $n$th root. I also tried to use a result: if $(x_n)\to x$ then $(\frac{x_n}{n})\to 1$ but don't know if this is true. Sadly I've had no real progress. Any help will be greatly appreciated, thanks in advance!
45932
Let$a_n,b_n\in\mathbb R$ and $(a_n+b_n)b_n\neq 0\quad \forall n\in \mathbb{N}$. The series $\sum_{n=1}^\infty\frac{a_n}{b_n} $ and $\sum_{n=1}^\infty(\frac{a_n}{b_n})^2 $ are convergent.  How to prove that$$\sum_{n=1}^\infty\frac{a_n}{b_n+a_n} $$ is convergent.  Thanks in advance
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?
64081
I have to show that the double serie $$\sum_{k=1}^{\infty} \sum_{i=1}^{\infty} ((\frac{1}{k+1} \cdot (\frac{k}{k+1})^{i}) - (\frac{1}{k+2} \cdot (\frac{k+1}{k+2})^{i}))$$ convergence to $-\frac{1}{2}$. Wolfram Alpha also says this series convergence to $-\frac{1}{2}$. Unfortunately, I don't know the calculation route.
82381
Is the following series convergent?  $$\sum_{n=1}^{\infty}\frac{e^n\,n!}{n^n}$$  I treid the Ratio and Root tests, but both of them failed.
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.
27086
$$\sum_{n=1}^\infty \frac{i^n}{n}$$  I'm not sure how to approach this series to find if it converges/diverges because of the $i^n$. I tried using the comparison test comparing it with $\frac{1}{n}$ and the ratio test, but didn't get anywhere. How would you approach this and look at series of this form or how to look at $i^n$ in an infinite series?
95818
$$\sum_{n=0}^\infty n!x^{n^2}$$   Can anyone tell me what will be the radius of convergence for this power series. According to me it is $0$.   If you can please show me how to go about it if I am wrong.
61361
Could I have a hint for testing the convergence of the following series please?  $$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$  Edit  The integral test does not work because $\int_1^n\frac{1}{(\ln x)^{\ln x}}dx$ has not an elementary primitive.  Thank You.
15494
Determine whether the series $\sum_{n=1}^{\infty}(\frac{n}{n+1})^{n^2}$ is convergent or divergent.  I've tried applying all the basic tests to no avail. I need to find out what the "trick" for this particular series is.
48067
I can clearly see that $\dfrac{n!}{n^n}\to 0$ when $n\to\infty$. But how do I know if the sum $$\sum_{n=1}^{\infty}\frac{n!}{n^n}$$ is convergent or not? I know this might be basic, but thank you if anyone can help me.
74043
Does the series $$ \sum_{n=1}^\infty \frac{\sin^2(n)}{n} $$   converge?   I've tried to apply some tests, and I don't know how to bound the general term, so I must have missed something. Thanks in advance.