friends.  The question is:   Find the value of $ \sum_{n=1}^{\infty} \frac{n^3}{3^n} $. I know this sum converges and that it's value is $ \frac{33}{8} $, however, I can't seem to find it.  I've tried doing $3S-S$ to try and find a pattern, tried using different subtractions, all to no avail.  Any help would be gladly accepted.  Thanks in advance, Pedro
124106
I was trying to solve $ \displaystyle \sum_{n = 1}^{\infty} \frac{n^3}{8^n}$ and I found a way to solve it and I want if there are generalizations for, say, $\displaystyle \sum_{n=1}^{\infty} \frac{n^k}{a^n}$ in terms of $k$ and $a$. I would also like to know if there is  a better way to solve it. Here's how I did it:  First I decomposed the series into the following sums:  $S_1  = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$  $S_2 = \frac{7}{64} + \frac{7}{512} + \dots = \frac{\frac{7}{64}}{\frac{7}{8}}$  $S_3 = \frac{19}{512} + \frac{19}{4096} + \dots = \frac{\frac{19}{512}}{\frac{7}{8}}$  And deduced that the sum can be written as $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{3n^2 - 3n + 1}{8^n}$  $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ is easy to evaluate -- it's $\frac{1}{7} $by geometric series  $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ can be evaluated in a whole host of ways to get an answer of $\frac{8}{49}$.  It remains to evaluate $\displaystyle \sum_{n = 1}^{\infty} \frac{n^2}{8^n}$, for which I took a similar approach as the cubics by decomposing it into many sums:  $T_1 = \frac{1}{8} + \frac{1}{64} + \dots = \frac{\frac{1}{8}}{\frac{7}{8}}$  $T_2 = \frac{3}{64} + \frac{3}{512} + \dots = \frac{\frac{3}{64}}{\frac{7}{8}}$  And so forth, coming to the conclusion that it is equal to $\frac{8}{7} \displaystyle \sum_{n = 1}^{\infty} \frac{2n-1}{8^n}$  Now, I used this information and the above values for $\displaystyle \sum_{n = 1}^{\infty} \frac{1}{8^n}$ and $\displaystyle \sum_{n = 1}^{\infty} \frac{n}{8^n}$ to get the sum as $\frac{776}{2401}$, which is confirmed by WA.  So, I would like to reiterate here: Is there a simpler way to compute this sum, and are there any known generalizations for this problem given an arbitrary $a$ in the denominator and arbitrary $k$ as the exponent in the numerator?
17492
Apparently,   $$ \sum_{n = 0}^\infty \frac{n}{2^n} $$  converges to 2. I'm trying to figure out why. I've tried viewing it as a geometric series, but it's not quite a geometric series since the numerator increases by 1 every term.
8268
I need to solve    $$\sum_{j = 1}^{\infty} \sqrt{\frac{j!}{j^j}}$$  Does this converge or diverge and why?
30910
Could I have a hint for testing the convergence of the following series please?  $$\sum_{n=2}^\infty\frac{1}{(\ln n)^{\ln n}}$$  I am very appreciative for your help.
27856
A previous problem had us solving $\sum_{n=0}^{\infty} \frac{1}{4^n}$ which I calculated to be $\frac{4}{3}$ using a bit of mathematical manipulation.   Wonderful. Thank you for all the prompt responses.  Could anyone suggest an alternate technique that does not involve differentiation?
81519
I just ran into an expression, and I would like to know what it converges to. $$\sum^{\infty}_{n=1} \frac{n}{(n-1)!}$$  Do you know if it converges to something (like $e$) or, in alternative, how to find out what it converges to?
36879
Does the following series converge:  $$\sum_{n=3}^\infty \frac {1}{(\log n)^{\log(\log(n)}}$$  I've tried all test I know... Any ideas ?
68289
Question is to check if : $\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}&gt;\frac{3}{5}$   the problem is that i am sure that this series$\sum_{n=1}^{\infty}\frac{(\log\log2)^n}{n!}$ is convergent (if it is not then i would be very happy as i will then be done.)  I would be thankful if someone can help me out with solving this.
92841
convergence of $ \sum_{k=1}^\infty \sin^2(\frac 1 k)$   How can I do this? Should I use the Ratio Test (I tried this but it started getting complicated so I stopped)? Or the Comparison test(what should I compare it to?)?
86275
I am struggling a bit with power series at the moment, and I don't quite understand what this question is asking me to do? Am I meant to form a power series from these, or simply evaluate that series? Any explanation/working is appreciated.  Using power series representation, calculate  $$\sum_{n=1}^\infty \frac{n2^n}{3^n}.$$
80108
I'd appreciate any help, I know it has something to do with the geometric series but I still can't figure out how. I thought about integration but couldn't find a way to do it. $$\sum_{n=1}^\infty \frac{1}{2^nn(3n-1)}$$  Thanks.
56816
$$ \sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}$$  Does it converge or diverge?   Can we have a rigorous proof that is not probabilistic?   For reference, this question is supposedly a mix of real analysis and calculus.
60322
Find the value of the following infinite series: $$\left\lfloor\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}\right\rfloor$$  Now, my doubt is whether it's​ $2$ or it's​ $3$. I'm not sure if it just converges to $3$ but not actually reaches it or if it completely attains the value of $3$.  I would to like to see the proof (formal) as well instead of just intuitions.
97009
Possible Duplicate:          please help me with this  $$\sum\limits_{k=0}^{\infty}\frac{k}{3^k}$$  I need just a hint, not a full answer. Thanks!
47326
How do you prove $$\sum_{n=1}^{\infty} \frac {n}{2^n} = 2\ ?$$  My attempt: I have been trying to find geometric series that converge to 2 which can bind the given series on either side. But I am unable to find these. Is there a general technique to find the sum? This is a high school interview question and must be easy enough to solve in a few minutes.  Please give any hints for the first step towards a solution.
44494
$$\sum_{n=1}^\infty \frac 1{n!}$$   I think we may compare it with $\frac{1}{n^n}$, and say that $\frac{1}{n!}&gt;\frac{1}{n^n}$. But I am not sure of the convergence or divergence of $\frac{1}{n^n}$ itself, so this turns out to be useless.  What other way is there?
119655
Determine when $\sum_1^\infty \frac{(2n)!x^n}{n(n!)^2}$ converges. By ratio test, when $|x|&lt;1/4$, the sum converges, $|x|&gt;1/4$ diverges. But I'm not sure about $|x|=1/4$. By Stirling's approximation $\frac{(2n)!(1/4)^n}{n(n!)^2}\sim\frac{4^n}{n{4^n}\sqrt{ \pi n}}\sim\frac{1}{\sqrt{\pi}n^{3/2}}$, so when $|x|=1/4$, the sum converges. But I haven't learned Stirling's approximation yet, so how to prove that when $|x|=1/4,$ the sum converges, without using it?