So all my friends keep telling me that if you add up all the numbers from 1 to infinity, (1+2+3+4...) then the answer is -1/12. They showed me this proof with infinite sums, but I didn't understand it, so all I'm asking is does it really equal -1/12?
1958
$\displaystyle\sum_{n=1}^\infty \frac{1}{n^s}$ only converges to $\zeta(s)$ if $\text{Re}(s) &gt; 1$.  Why should analytically continuing to $\zeta(-1)$ give the right answer?
97514
I am trying to sum the following series, given n is a positive integer, $$ n+n(n-1)+n(n-1)(n-2)+...+n! $$ I think there is a solution, but if not, an approximate result will also be appreciated.   Regards.
12710
Problem :   How to find the sum of the sequence $\frac{1}{1+1^2+1^4} +\frac{2}{1+2^2+2^4} +\frac{3}{1+3^2+3^4}+.....$  I am unable to find out how to proceed in this problem.. this is a problem of arithmetic progression... Please suggest how to proceed...Thanks..
8533
Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$    I think this is known (see ), I appreciate any hint or link for the solution (or the full solution).
28191
I'm sure there's an identity or some known "trick" to solve this but I couldn't find it (Looked in Wikipedia, Wolfram and here):  $$ \sum_{k=0}^n (-1)^k{n \choose k} =0 $$  I want to know how this can be proven. I tried induction but I'm not sure what to do with ${{n+1}\choose k}$.  Thanks!
41502
Calculate  $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$, I do not know hot get rid of that $k$, for me it is similar like  $\binom{n}{k}=\frac{k}{n} \binom{n-1}{k-1}$, do you have some idea?
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!
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.
40271
I've got the following sum which I'm trying to figure out:  $$\sum_{x=0}^n 2^{-x}$$  Wolframalpha tells me that it's equal to $2 - 2^{-n}$ but I am interested in figuring out why, and how to get that result by hand.  Any help is greatly appreciated. Thanks.  Also, as you can tell, I could really use some help with MathJax...
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?
38666
$\sum\limits_{n=1}^{\infty}\frac{1}{1+z^n}$, $|z|&gt;1$.  There are two facts that my professor uses that I am confused about.  The first is: $|1+z^n| \geq ||z|^n-1|$, I believe this is true for any $|z|$.  The other is: $\frac{1}{|z|^n-1} \leq \frac{2}{|z|^n}$, I believe this is also true for any $|z|$.  Can anyone prove these statements for me?
17936
Find the value of $\sum_0^n \binom{n}{k} (-1)^k \frac{1}{k+1}$.  Writing out several terms, I think the answer is $1/(n+1)$, but I'm struggling to prove this. I would greatly appreciate it if someone could show how to prove this.
45648
Since everyone freaked out, I made the variables are the same. $$ \sum_{x=1}^{n} 2^{x-1} $$  I've been trying to find this for a while. I tried the usually geometric equation  but I couldn't get it right (if you need me to post my work I will). Here's the outputs I need:  1, 3, 7, 15, 31, 63  If my math is correct.
33878
Can sum of a rational number and its reciprocal be an integer?   My brother asked me this question and I was unable to answer it.   The only trivial solutions which I could think of are $1$ and $-1$.  As to what I tried, I am afraid not much. I have never tried to solve such a question, and if someone could point me in the right direction, maybe I could complete it on my own.  Please don't misunderstand my question.   I am looking for a rational number $r$ where $r + \frac{1}{r}$ is an integer.
5055
I was messing around with some math formulas today and came up with a result that I found pretty neat, and I would appreciate it if anyone could explain it to me.   The formula for an infinite arithmetic sum is $$\sum_{i=1}^{n}a_i=\frac{n(a_1+a_n)}{2},$$ so if you want to find the sum of the natural numbers from $1$ to $n$, this equation becomes $$\frac{n^2+n}{2},$$ and the roots of this quadratic are at $n=-1$ and $0$. What I find really interesting is that $$\int_{-1}^0 \frac{n^2+n}{2}dn=-\frac{1}{12}$$ There are a lot of people who claim that the sum of all natural numbers is $-\frac{1}{12}$, so I was wondering if this result is a complete coincidence or if there's something else to glean from it.
28253
While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\sum_{n=2}^{\infty} \left(\zeta(n)-1\right) = 1 $$ Furthermore, if you use only the even numbers for the zeta function, the sum will converge to $\frac{3}{4}$, or $$\sum_{n=1}^{\infty} \left(\zeta(2n)-1\right) = \frac{3}{4}$$  Leaving $$\sum_{n=2}^{\infty} \left(\zeta(2n-1)-1 \right)= \frac{1}{4}$$  This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so?
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?