So i have this sum:  $$\sum_{n=2} ^{\infty} \frac{1} {n^3-n} $$  So for now i tried using this:  $$\sum_{n=2} ^{\infty} \frac{1} {n^3-n} =  \frac{1} {6}(1+\frac{1} {4}+\frac{1} {10}+...$$  Since $n^3-n=n(n-1)(n+1) $ is divisible by 6.(The multiplication of k consecutive numbers are always divisible by k! )  But i am stuck as i dont identify any known series in here.   Any help eould be appreciated.
74621
As my question says,   $$\frac 1{1.2.3}+\frac 1{2.3.4}+\frac 1{3.4.5}+...$$  As far as $\frac 1{1.2}+\frac 1{2.3}+\frac 1{3.4}+...$ is concerned, we can write it as $\frac 1{n-1} - \frac 1{n}$ What can we do here?
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
20763
Trying to find the sum of the following infinite series:  $$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$  Any ideas on how to find this sum?
79
How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$  I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty  \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1}  \frac{1}{m^2+(N-m)^2}$$  $$\frac{1}{m^2+(N-m)^2}\leq \frac{2}{N^2}$$ but it doesn't work.
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.
9714
Find this sum $$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}$$  My try: since $$n^4+n^2+1=(n^2+1)^2-n^2=(n^2+n+1)(n^2-n+1)$$ then $$\sum_{n=1}^{\infty}\dfrac{1}{n!(n^4+n^2+1)}=\sum_{n=1}^{\infty}\dfrac{1}{2n\cdot n!}\left(\dfrac{1}{n(n-1)+1}-\dfrac{1}{n(n+1)+1}\right)$$ then I can't go further.
122727
I need to show that $\sum (-1)^{n+1} / n$ = $k$ (which is $1 - 1/2 + 1/3 - 1/4+...$) can be expanded to a sum with $ 1 - 1/2 + 1/3 + 1/5 - 1/4 +1/7 + 1/9 - 1/6 + .... $ which is $3k/2$.   How can I start? I know that $k/2$ = $\sum (-1)^{n+1}/2n$ How can I go further so I get $3k/2$?
13007
What is the sum of $$\sum_{n=0}^{\infty}\frac{1}{2^n(n+1)}$$    I've spent an insane amount of time on this problem. I checked on Wolfram and it gives $ln(4)$, which I assume you get from $2\ln(2)$.   I know that the integral of $\frac{1}{n+1}$ is $\ln(n+1)$. What am I missing?  Thank you!
91606
I'm supposed to determine whether this sum diverges or converges and if it converges then find its value:  $$ \sum_{n=2}^\infty \frac{1}{n^2-1}. $$  Using the comparison test I eventually showed that this converges.  But I can't figure out how to show what this sum converges to.  The only sums we actually found values for in my notes are geometric series which this clearly isn't.  I saw that I could use partial fraction decomposition to represent the terms as  $$\frac{1/2}{n-1}- \frac{1/2}{n+1} $$ but that just gets me $\infty - \infty$, so this isn't the way to do it.  I'm not sure how to find the value of this sum.  I don't need the full solution but a hint would be appreciated. Thanks. :)
39742
I have a question regarding the sum of this series: $$\frac{3}{1!+2!+3!} + \frac{4}{2!+3!+4!} + \cdots + \frac{2008}{2006!+2007!+2008!}$$ My approach: I found that this sum is equal to: $$\sum_{n=3}^{2008}\frac{n}{(n-2)!+(n-1)!+(n)!}$$  I reduced it to : $$\sum_{n=3}^{2008}\frac{1}{n(n-2)!}$$  Please suggest how to proceed further.
7372
I'm not sure if I should call this "the sum of a sum" but I can't think of another phrase for it.   Anyhow, I'm confused about excercises such as this one  $$\sum_{j=2}^{\infty} \left( \sum_{k=2}^{\infty} k^{-j} \right)$$  where I want to compute the sum. I'm thinking maybe I could rewrite it as  $$\sum_{k=2}^{\infty} \frac{1}{k^{2}} + \sum_{k=2}^{\infty} \frac{1}{k^{3}}+... $$  but I don't know where to go from there or if that is even correct thinking. (I just began my first course in linear analysis.)
74045
Let $$\displaystyle a_n=\sum_{n=2}^{\infty} \frac{(-1)^n}{n^{\frac{1}{3}}+(-1)^{\frac{n(n+1)}{2}}}$$   so I divide it into four series $4k, 4k+1, 4k+2, 4k+3$ and I pair for instance $4k, 4k+1$ and $4k+2, 4k+3$ and prove that these two series is convergent and conclude that since  both series are convergent so the sum of it is also convergent but I'm not sure if it's legall   second series I'm not sure is like this $\displaystyle b_n=\frac{1}{1}+\frac{1}{3}-\frac{1}{2}+\frac{1}{5}+\frac{1}{7}-\frac{1}{4}+\frac{1}{9}+\frac{1}{11}-\frac{1}{6}+...$   here I also split it into smaller parts namely $\displaystyle \sum_{n=1}^{\infty}\frac{1}{4n-3}+\sum_{n=1}^{\infty}\frac{1}{4n-1}-\sum_{n=1}^{\infty}\frac{1}{2n}$ and I prove that $\displaystyle \frac{1}{4n-3}+\frac{1}{4n-1}-\frac{1}{2n}&lt;\frac{1}{n^2} \cdot c $ for some const c so it implies that origin series is convergent or not ?
26706
Suppose $\sum_{n=1}^\infty \frac{1}{n} = S$ where $S$ is finite.  Then  $$S =\sum_{n=1}^\infty \frac{1}{n}= \sum_{n=1}^\infty \frac{1}{2n-1} + \frac{1}{2n} &gt; \sum_{n=1}^\infty \frac{1}{2n} + \frac{1}{2n} = S$$  which is a contradiction.  Is this valid?
49817
Lets define the following series $$S_n =\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}.$$ I know that $S_0$ does not converge so let's suppose $n \in N$ and define $S_n$ for some $n$. We have $S_1=1$ , $S_2=\frac1 4$ , $S_3=\frac 1 {18}$ , $S_4=\frac 1{96}$ , $S_5=\frac 1{600}$ etc.. the numerator in all results is 1the pattern in denominator is $[1,4,1,96,600...]$ and   that is equal to $n*n!$ Finally I want to prove the general equality: $\sum_{i=1}^{\infty} \frac 1 {i(i+1)(i+2)...(i+n)}=\frac 1 {n*n!}$
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.
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  $$\sum_{n=1}^\infty \frac{n2^n}{3^n}.$$
29310
Find the sum to $n$ terms as well as the sum to infinity of the series:  $$\frac{1}{3} + \frac{1}{4\cdot 2!} + \frac{1}{5\cdot 3!} +\cdots.$$  I was trying this question many times but I didn't get the answer. I was using geometric series formula, but it was not applicable here...I don't know where I have to start, and I don't know any hint about this series....  If anybody helps me, I would be very thankful to him.