When I learned about how the sum of a series could be divergent and convergent, I decided to try to find the sum of the square reciprocals to no avail. The proof, obviously at not my level as you can tell, was too difficult to understand (what math branch is required because they mentioned something a Riemann Zeta function).  Is there an alternative proof that would yield the result ${\pi^2}/6$ as well without higher math?
24765
As I have heard people did not trust Euler when he first discovered the formula (solution of the ) $$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}.$$ However, Euler was Euler and he gave other proofs.   I believe many of you know some nice proofs of this, can you please share it with us?
36066
So I need to know the convergence of this sum. The integral test does not seem to work (WolframAlpha gives a confusing answer which implies it's not the correct method). I also tried the limit comparison test for various harmonic series, but they are all inconclusive. I'm also not getting very far with the ratio test. $$ \sum_{n=2}^\infty \dfrac{\sqrt{n+1}}{n(n-1)} $$
45732
Does this series converge or diverge?  $$\sum_{n=1}^\infty\frac{\sqrt{n+1}-\sqrt{n}}{n}$$  I tried the comparsion test with $\sum_{n=1}^\infty$ $1 \over n$ but this does not help.  I also tried to simplify the series to $\sum_{n=1}^\infty \frac{1}{n (\sqrt{n+1}+\sqrt{n}{}{}{})}$ but this become harder.
44542
How can one prove that the series $\displaystyle \sum\limits_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number? There's no need to use Taylor expansion, integrals or any advanced/professional techniques. It can be proved using only basic techniques. My first attempt was to assume that the series does converge to a rational number $ \frac{p}{q}$ and then it follows by definition that for every positive epsilon there exists an integer k such that $ {\sum_{n=1}^{k} \frac{1}{(n!)^2} -\frac{p}{q}}&lt;=\epsilon$ I tried to get a contradiction but I failed.
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.)
60070
How can we evaluate this series  $$\sum_{n=1}^\infty \frac{n^a}{b^n}?$$  Here $a$ and $b$ are positive integers. If $b=1$ then series will be diverging, in other cases, it will be converging, but how to find this sum?  Isn't there a simple solution which doesn't involve Stirling numbers?
97412
I'm looking for a way to prove $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{(-1)^m E_{2m}\pi^{2m+1}}{4^{m+1}(2m)!}$$ I know that $$\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{1}{4^{2m+1}}\left(\zeta\left(2m+1,\frac14\right)-\zeta\left(2m+1,\frac34\right)\right)$$ so maybe I could simplify the above more?
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?
100523
Why is it that  make sense?  Specifically, by basic calculus a sum such as $1 - 1 + 1 ...$ describes a divergent series (where divergent := non-convergent sequence of partial sums) but, as described , one can use Euler, Borel or generic summation to arrive at a value of $\tfrac{1}{2}$ for this sum.  The first apparent indication that this makes any sense is the claim that the summation machine 'really' works in the complex plane, so that, for a sum like $1 + 2 + 4 + 8 + ... = -1$ there is some process like this:    going on on a unit circle in the complex plane, where those lines will go all the way back to $-1$ explaining why it gives that sum.   The claim seems to be that when we have a divergent series it is a non-convergent way of representing a function, thus we just need to change the way we express it, e.g. in the way one . This claim doesn't make any sense out of the picture above, &amp; I don't see (or can't find or think of) any justification for believing it.  Futhermore there is this notion of  one uses in Fourier theory. For some reason one can construct these Cesaro sums to get convergence when you have a divergent Fourier series &amp; prove some form of convergence, where in the world does such a notion come from? It just seems as though you are defining something to work when it doesn't, obviously I'm missing something.  I've really tried to find some answers to these questions but I can't. Typical of the explanations is  of Hardy's divergent series book, just plowing right ahead without explaining or justifying the concepts.  I really need some general intuition for these things for beginning to work with perturbation series expansions in quantum mechanics &amp; quantum field theory, finding 'the real' expanation for WKB theory etc. It would be so great if somebody could just say something that links all these threads together.
57775
What is the value of  $$\sum_{n=1}^{\infty} \frac{n}{2^{\sqrt{n}}}$$  It's clearly convergent but is it possible to calculate the sum?
120838
Suppose the infinite series $\sum a_n$ converges ($a_n \in \mathbb{R}$ not all positive or negative).  (1) Is it true that $\sum a_n^2$ converges?  (2) Is it true that $\sum a_n^3$ converges?  For (1), the answer is no. I found the counterexample $a_n = (-1)^n / \sqrt{n}$ which converges by alternating series, but $\sum a_n^2 = \sum1/n $ is divergent.  I'm unsure about proving truth of (2) and can't seem to find a counterexample.  Thanks for help.
16604
Apery's original proof that $$ \zeta(3) \equiv \sum_1^\infty \frac1{n^3} $$ is irrational starts from an alternating series  $$ \zeta(3) = \frac52\sum_1^\infty\frac{(-1)^{n-1}}{n^3\binom{2n}{n}} $$  There must be a way to see that those two series are equal, but I have tried various telescoping and other techniques and I just can't see it. Help me out:       Show (preferably by reasonably elementary means) that$$\sum_1^\infty \frac1{n^3}=\frac52\sum_1^\infty\frac{(-1)^{n-1}}{n^3\binom{2n}{n}}$$
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?
12931
$$(\sqrt 2-\sqrt 1)+(\sqrt 3-\sqrt 2)+(\sqrt 4-\sqrt 3)+(\sqrt 5-\sqrt 4)…$$  I have found partial sums equal to natural numbers. The first 3 addends sum to 1. The first 8 sum to 2. The first 15 sum to 3. When the minuend in an addend is the square root of a perfect square, the partial sum is a natural number. So I believe this series to be divergent.  Am I right? Have I used correct terminology? How would this be expressed using sigma notation? Is there a proof that this series diverges?
80975
Study the convergence of $$\displaystyle \sum_{n\geq 1} \frac{(-1)^n|\sin n|}{n}.$$   I am stuck with this series, we need probably some measure of irrationally of $\pi$, unfortunately I am unfamiliar with this. So here is my attempt :  Let $f(x) = \sum \frac{|\sin{n}|}{n} x^n, |x| &lt; 1$  It's not difficult to compute the Fourier series of $|\sin(x)|$ :  $$ \displaystyle|\sin(x)|=\frac{2}{\pi}-\frac{4}{\pi}\sum_{n=1}^{+\infty}\frac{\cos(2nx)}{4n^2-1} $$  Then Fubini's theorem () works very well (because the previous series converges absolutely at $x$ fixed ) and all calculations made, we find that for all $x\in( -1,1)$:  $$ \displaystyle f(x)=\frac{2}{\pi}\sum_{n=1}^{+\infty}\frac{x^n}{n}-\frac{4}{\pi}\sum_{p=1}^{+\infty}\frac{x^2-2x\cos(p)}{(4p^2-1)(x^2-2x\cos(p)+1)} $$  However, the second sum I have not been able to show the convergence. I feel the series diverge because the following series $$ \displaystyle\sum\frac{1}{p^2\sin^2\left(\frac{p}{2}\right)} $$  diverge because  $0$ is an accumulation point of $\displaystyle (n\sin(n))$ sequence.  Any ideas (for the original series) ?
68154
The sum of the series $$ \frac{\pi}{2}=\sum_{k=0}^\infty\frac{k!}{(2k+1)!!}\tag{1} $$ can be derived by accelerating the  $$ \frac{\pi}{4}=\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\tag{2} $$ using . Mathematica is able to sum $(1)$, so I assume there must be some method to sum the series in $(1)$ directly; what might that method be?
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.