While watching interesting mathematics videos, I found one of the papers  of Srinivasa Ramanujan to G.H.Hardy in which he had written $1^3+2^3+3^3+4^3+\cdots=\frac{1}{120}$.  The problem is that every term on the left is more than $\frac{1}{120}$ yet the sum is $\frac{1}{120}$. How is that ???  I know that there are many and much more interesting things  presented by Ramanujan (like $1-1+1-1+1...=\frac{1}{2}$ and $1+2+3+4.....=\frac{-1}{12}$) but for now I am interested in the summation in the title. Any idea/hint is heartily welcome. Thanks.   the video I'm talking about.
121052
It seems that under the light of  the following is plausible:  $$1 + {2^{2n - 1}} + {3^{2n - 1}} +  \cdots  =  - \frac{{{B_{2n}}}}{{2n}}(\Re)$$  Alas, I can't really find any concrete definition of Ramanujan Summation. Could someone provide a small explanation?
64621
Can anyone tell me about the sum of the series $$\sum_{n=1}^\infty \frac{1}{(2n)(2n+1)}?$$ This is not a usual telescoping sum in which all the terms cancel out.
77605
A closed form for the sum $S = \frac {2}{3+1} + \frac {2^2}{3^2+1} + \cdots + \frac {2^{n+1}}{3^{2^n}+1}$ is $1 - \frac{a^{n+b}}{3^{2^{n+c}}-1}$, where $a$, $b$, and $c$ are integers. Find $a+b+c$.
103037
Prove that $$1+\frac{1}{3}+\frac{1\cdot 3}{3\cdot 6}+\frac{1\cdot 3\cdot 5}{3\cdot 6 \cdot 9}+.........=\sqrt{3}$$   $\bf{My\; Try::}$ Using Binomial expansion of  $$(1-x)^{-n} = 1+nx+\frac{n(n+1)}{2}x^2+\frac{n(n+1)(n+2)}{6}x^3+.......$$  So we get $$nx=\frac{1}{3}$$ and $$\frac{nx(nx+x)}{2}=\frac{1}{3}\cdot \frac{3}{6}$$  We get $$\frac{1}{3}\left(\frac{1}{3}+x\right)=\frac{1}{3}\Rightarrow x=\frac{2}{3}$$  So we get $$n=\frac{1}{2}$$  So our series sum is $$(1-x)^{-n} = \left(1-\frac{2}{3}\right)^{-\frac{1}{2}} = \sqrt{3}$$  Although I know that this is the simplest proof, can we solve it any other way someting like defining $a_{n}$ and then use Telescopic sum.  Thanks.
49156
In an , I stated $$\sum\limits_{j=0}^M \frac{M \choose j}{N+M \choose j} = \frac{N+M+1}{N+1}.$$  It is clearly true when $N=0$ since you add up $M+1$ copies of $1$, and when $M=0$ since you add up one copy of $1$.  And, for example, with $M=4$ and $N=9$ you get $\frac{1}{1}+\frac{4}{13}+\frac{6}{78}+\frac{4}{286}+\frac{1}{715} = \frac{14}{10}$ as expected.  But how might you approach a general proof?
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?
100751
How do I find the formula for this sum  $S_n = \sum_{i=1}^n i(i-1)$  Wolfram alpha gives me the correct formula which is $ \frac{1}{3} (n-1)n(n+1)$, but I'm interested in how I get to that and how I can approach this kind of problem in general.
10232
I came across a question where I needed to find the sum of the factorials of the first $n$ numbers. So I was wondering if there is any generic formula for this? Like there is a generic formula for the series: $$ 1 + 2 + 3 + 4 + \cdots + n = \frac{n(n+1)}{2} $$ or $$ 1^{2} + 2^{2} + 3^{2} + 4^{2} + \cdots + n^{2} = \frac{n(n+1)(2n + 1)}{6} $$  Is there is any formula for: $$ 1! +2! +3! + 4! + \cdots + n! $$ and $$ {1!}^2 +{2!}^2 +{3!}^2 + \cdots + {n!}^2 $$?  Thanks in advance. If not, is there any research on making this type of formula? Because I am interested.
40584
Find this sum     $$I=\sum_{n=0}^{\infty}\dfrac{1}{(3n+1)(3n+2)(3n+3)}$$   My try: let     $$f(x)=\sum_{n=0}^{\infty}\dfrac{x^{3n+3}}{(3n+1)(3n+2)(3n+3)},|x|\le 1$$   then we have $$f^{(3)}(x)=\sum_{n=0}^{\infty}x^{3n}=\dfrac{1}{1-x^3}$$ then we find the $f(x)$,Following is very ugly(can you someone can post your follow solution,) have other simple methods? Thank you very much.
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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.
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 $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
41068
Problem is to find sum  $$\frac{1}{3!}+\frac{4}{4!}+\frac{9}{5!}\cdots$$  What I knew doesn't apply on this problem  Some series are telescoping, some types are solvable using binomial , both look useless here Binomial gives  $$n (n-1) (n-2)x^3=1$$  $$n (n-1)( n-2)(n-3)x^4=4$$  What approach to use here?
54083
The sum came up when I was trying to evaluate $\frac{1}{1}+\frac{1}{11}+\frac{1}{11}\cdots$  I rewrote the sum as, $S = 9 \ \displaystyle{\sum_{i=1}^n} \frac{1}{10^i-1}$  Wolfram gives an answer in terms of the digamma function.
52987
I'm currently a Combinatorics student trying to parse through this solution. I do not understand the proof currently. Any help understanding it is greatly appreciated.    Question Let the number of partitions be given by the function $p(n)$, i.e. the number of ways of writing the integer $n$ as a sum of positive integers. Prove that that if $g$ is any polynomial function in $n$, then there exists an integer $N$ such that $g(n)&lt;p(n)$ for all $n&gt;N$. Do not use the asymptotic Hardy-Ramanujan formula.    Provided Proof.    Enough to show $n^a&lt;p(n)$ for fixed positive integer $a$. (This is just saying that we want to show $p(n)$ grows faster than any degree polynomial of $n$, right?) We have that $(i_1,i_2,\ldots,i_{a+1},1^{n-i_1-\cdots -i_{a+1}})$ is a partition of $n$ as long as $i_{a+1}\leq i_a\leq \cdots i_1 \leq \lfloor\frac{n}{a+1}\rfloor$. (I really don't understand this step. What is this partition? Of the coefficients or terms of polynomial $p(n)$? And how do we know this is $\leq \lfloor\frac{n}{a+1}\rfloor$?) The number of partitions $n$ of this form is multiset binomial coefficient $$\left(\dbinom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)=\frac{\lfloor\frac{n}{a+1}\rfloor(\lfloor\frac{n}{a+1}\rfloor+1)\cdots(\lfloor\frac{n}{a+1}\rfloor+a-1)}{(a+1)!}$$ (I'm assuming that if I can understand (2.) then I can understand this line) This means $p(n)&gt;\left(\binom{\lfloor \frac{n}{a+1}\rfloor}{a+1}\right)$ for all $n&gt;0$. Since $a$ is constant, the given fraction grows with order $n^{a+1}$, which means $p(n)&gt;n^a$ for $n$ sufficiently large. (Why does it grow with order $n^{a+1}$?)   Thanks all again in advance!
42038
Mathematica tells me that $\sum\limits_{i=1}^n \frac1{2i-1}$ is equal to $\frac12 H_{n-1/2}+\log\,2$, where $H_n$ is a harmonic number.  Why is this true? Is there a general strategy for evaluating sums of the form $\sum\limits_{i=1}^n \frac1{ai+b}$?
28610
Find $$ \sum_{r=1}^n \frac{2r+3}{3^r (r+1)} $$   I don't understand how to split this fraction into partial fractions which I need to do to complete the rest of the question
50237
I know that $$1^3+2^3+3^3+\cdots+n^3=\left(\frac{n(n+1)}{2}\right)^2$$  What I would like to know is whether there is a simple proof (that obviously does not use the above info) as to why the sum of the first $n$ cubes always a perfect square.