The staff of  has shown that the sum of all the integers from $0$ to $\infty$ is $-\frac1{12}$. Recently I was looking for the sum of all the (positive) integers from $0$ to $n$ and I found that: $$\sum_{i=0}^n i=\frac{n(n+1)}{2}$$  So I decided to take the limit: $$\lim_{n\to \infty}\frac{n(n+1)}{2}$$ but that tends towards $\infty$ when I expected that to be $-\frac1{12}$! Where did I got wrong? (the result is also confirmed by Wolfram Alpha)
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?
122723
I'm noticing some things:  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/3}-\frac{3}{2}n^{2/3} \right)=\zeta(1/3)$$  Note $\int n^{-1/3} dn=\frac{3}{2}n^{2/3}+c$  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/2}-\frac{2}{1}n^{1/2} \right)=\zeta(1/2)$$  And  $$\int n^{-1/2}dn=2n^{1/2}+c$$  It seems as though  $$\lim_{n \to \infty} \left(\sum_{x=1}^{n} x^{-1/s}-\frac{s}{s-1}n^{(s-1)/s} \right)=\zeta(1/s)$$  If $s \neq 1$, may someone please explain why.
100131
I want to calculate the sum:$$\sum_{n=1}^\infty\frac{1}{(n-1)!(n+1)}=$$  $$\sum_{n=1}^\infty\frac{n}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{n+1-1}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{n+1}{(n+1)!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}=$$ $$\sum_{n=1}^\infty\frac{1}{n!}-\sum_{n=1}^\infty\frac{1}{(n+1)!}.$$  I know that $$\sum_{n=0}^\infty\frac{1}{n!}=e$$  so $$\sum_{n=1}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{n!}-1=e-1$$  But, what can I do for $$\sum_{n=1}^\infty\frac{1}{(n+1)!}$$ ?  Am I allowed to start a sum for $n=-1$ ? How can I bring to a something similar to $$\sum_{n=0}^\infty\frac{1}{n!}$$?
114880
We have the sum:  $\sum\nolimits_{n=0}^\infty \frac{n}{n!}$  Why is this sum a representation of $e$ when the sum:  $\sum\nolimits_{n=0}^\infty \frac{1}{n!}$  is already a representation of $e$?  For me it seams that the first row is greater then the second.
73042
Here's the problem: Let $x$ be any real number. Show that $$ \lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \sum_{n=0}^ \infty \frac{x^n}{n!}  $$ I'm sure there are many ways of pulling this off, but there are 3 very important hints to complete the exercise in the desired manner:   Expand the left side as a finite sum using the Binomial Theorem. Call the summation variable $n$. Now add into the finite sum extra terms which are $0$ for $n&gt;m$, in order to make it look like an infinite series. What happens to the limit on $m$ outside the series?   So far I was able to use Hint 1 to expand the left side: $$ \lim_{m \to \infty} \left( 1 + \frac{x}{m} \right)^m = \lim_{m \to \infty} \sum_{n=0}^m \binom {m}{n} \left( \frac{x}{m} \right)^n $$  No matter what I do with the binomial coefficients and factorials, I can't figure out what extra terms to add per Hint 2. Any suggestions?
92775
Let $\alpha$ be a positive number. Calculate $$\lim_{n\to\infty} \sum_{k=1}^{n} \Big(\frac{k}{n}\Big)^{\alpha k}$$  Edit: I have deleted my attempt, it didn't seem to lead me anywhere and I discovered a couple of mistakes. :)  Please give me a hint!
33882
I have been trying around different ranges of summations:  $$ \sum_{(m,n) \in \mathbb{Z}^2} \frac{m^2 - n^2}{(m^2 + n^2)^2} = 0$$  That's not any good.  What about if we restrict to $m, n \in \mathbb{Z}$ as positive integers.    $$ \sum_{m &gt; 0, n &gt; 0} \frac{m^2 - n^2}{(m^2 + n^2)^2} = 0$$  Now here's an anti-symmetry.  I do not like taking out the symmetry, but perhaps I can ask about:   $$ \sum_{m &gt;  n &gt; 0} \frac{m^2 - n^2}{(m^2 + n^2)^2}  = \; ? \tag{$*$} $$  There doesn't seem to be a change of variables that can work.  And we've used symmetry about as much as we can.  This looks related to:  $$  \sum \frac{1}{n^2} = \frac{\pi^2}{6}$$  Perhaps this other series ($*$) also has a special value.
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?
855
If $n&gt;1$ is an integer, then $\sum \limits_{k=1}^n \frac1k$ is not an integer.  If you know , then you know there must be a prime $p$ between $n/2$ and $n$, so $\frac 1p$ appears in the sum, but $\frac{1}{2p}$ does not. Aside from $\frac 1p$, every other term $\frac 1k$ has $k$ divisible only by primes smaller than $p$. We can combine all those terms to get $\sum_{k=1}^n\frac 1k = \frac 1p + \frac ab$, where $b$ is not divisible by $p$. If this were an integer, then (multiplying by $b$) $\frac bp +a$ would also be an integer, which it isn't since $b$ isn't divisible by $p$.  Does anybody know an elementary proof of this which doesn't rely on Bertrand's Postulate? For a while, I was convinced I'd seen one, but now I'm starting to suspect whatever argument I saw was wrong.
84235
I'm asked to prove that $$\lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)=\frac{\pi}{4}$$ This looks like it can be solved with Riemann sums, so I proceed:  \begin{align*} \lim_{n \to \infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+4}+\frac{n}{n^2+9}+\cdots+\frac{n}{n^2+n^2}\right)&amp;=\lim_{n \to \infty} \sum_{k=1}^{n}\frac{n}{n^2+k^2}\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{n^2}{n^2+k^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}(\frac{1}{n})(\frac{1}{1+(k/n)^2})\\ &amp;=\lim_{n \to \infty} \sum_{k=1}^{n}f(\frac{k}{n})(\frac{k-(k-1)}{n})\\ &amp;=\int_{0}^{1}\frac{1}{1+x^2}dx=\frac{\pi}{4} \end{align*}  where $f(x)=\frac{1}{1+x^2}$. Is this correct, are there any steps where I am not clear?
5154
What is the result of the next limit:   $$ \lim_{n \to \infty} \frac{ \sum^n_{i=1} i^k}{n^{k+1}},\ k \in \mathbb{R} $$  Why (theorem)?
28250
Consider $$\sum_{n=1}^{\infty}\frac{(-1)^{\pi(n)}}{n}$$ where $\pi(n)$ is the number of primes less than or equal to $n$.  Does this sum converge or does it diverge? Are there any results related to this?
114099
Given the Series $$\sum_{k=1}^\infty \frac{1}{k(k+2)}$$  How exactly would I find out the limit is $\frac34$ as suggested by Wolframalpha? I already found out I can prove it actually converges by performing the comparison test and seeing that the underlying sequence isn't a null-sequence. But unfortunately I am absolutely clueless on how to prove that it converges to $\frac34$.  Regards,  Dennis
93825
$\lim \limits_{x \to 1^-} \displaystyle \sum_{n=0}^\infty (-1)^nx^{n²} = \frac{1}{2}$  The power $n^2$ is problematic. Can we bring this back to the study of usual power series?  I do not really have any idea for the moment.
103605
Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[  \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$  I have a question about the following solution:  We may write it in the form:  $$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$  Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 &lt; \frac{1}{n} &lt; \ldots &lt; \frac{n}{n}$.  Can you explain to me to reach this conclusion?
16311
As seen in  question, the series $\sum\limits_{n=1}^{\infty} \dfrac{n!e^n}{n^n}$ diverges. (One way to see this is by noting that the terms of the sum are greater than $1$ and, therefore, don't converge to zero.)  However, more is true: not only are the terms greater than $1$, they blow up to $+\infty$! To see this, one can apply Stirling's approximation and get $\dfrac{n!e^n}{n^n} \sim \sqrt{2\pi n}$.  My question is: is there an elementary proof for $\lim_{n \to\infty}\dfrac{n!e^n}{n^n} = +\infty$?  The reason I'm asking this is that I'm taking a Calculus course and I was assigned the exercise of proving the divergence of the series above and I would feel more satisfied if I could prove that its terms go to $+\infty$ (as opposed to merely proving the terms are greater than $1$.) Obviously, things like Stirling's approximation are outside the scope of the course, so I can't really use them.  (To be clear, you can use anything one learns in $4$ semesters of standard Calculus courses. I hope this restriction is not too obscure.)
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!