I'm a high-school senior attempting to make sense of the zeta function. I know Riemann regularized it to include complex numbers. Apparently, from this we could obtain that the sum of natural numbers may be assigned the value '-1/12' (zeta of -1)  What I don't get is how did extending the domain to complex numbers help in getting this result.  Thanks,
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?
52365
Background: This is a step from a longer proof/exercise that  $\sum_{n=1}^{\infty}\tau (n)^2/n^s=\zeta(s)^4/\zeta(2s)$ for $\sigma&gt;1$  Expanding the sum and using counting I get:   $\frac{1+t}{(1-t)^3}=(1+t)(1+{3\choose 1}t+\left ({3\choose 1}+{3\choose 2}\right )t+\left({3 \choose1}+3!+1\right )t^2 \ldots=\sum_{n=0}^{\infty}(1+n)^2t^n$$  So the final coefficients are each coefficient in the infinite sum plus the coefficient before it because of the $(1+t)$ This gives 1, 4, 9, 16 How would one calculate explicitly (as the author likely did) that the coefficients of $\frac{1+t}{(1-t)^3}$ are $(n+1)^2$ rather than counting and recognizing a pattern?
81871
I recall from complex analysis a while ago that the series $$\sum_{n=1}^\infty \frac{z^{n!}}{n}$$ has radius of convergence 1, diverges on roots of unity, and converges elsewhere on the circle. It's simple enough to see that it has to diverge on roots of unity: consider $z$ such that $z^k = 1$ for some $k \in \mathbb{N}.$ Plug this into the series and you see that past the $k$'th term, $z^{n!} = (z^k)^m = 1$ for some $m \in \mathbb{N}$; hence, the series becomes harmonic and must diverge.  But I don't recall how to show that the series converges for other values on the circle. If this is simple I'd like just a hint, if it is involved I would prefer a sparse outline of the proof. Thank you!
28253
While I was looking at the values of the zeta function for the first natural numbers, I noticed that the sum of the values minus $1$, converge to $1$. Better put: $$\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}$$  This is probably common knowledge among mathematicians, but I couldn't find much about it on the internet. Is there a proof of this or perhaps even a simple explanation why this is so?
80843
I've been reading a research paper, and I'm interested in generalizing a certain theorem but I can't seem to understand how the following results are derived:  $$\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{1}{z_i-z_j}=0$$  $$\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i}{z_i-z_j}=\frac{1}{2}n(n-1)$$ $$\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i^2}{z_i-z_j}=(n-1)\sum_{i=1}^n z_i$$ $$\sum_{i=1}^{n}\sum_{j\neq i}^n \frac{z_i^3}{z_i-z_j}=(n-1)\sum_{i=1}^n z_i^2 + \sum_{i&lt;j}^n z_i z_j$$
60973
I tried my best to solve this problem from what I learned in residues, but the solution seems very far from what I was doing!! Is there any way other than using Laurent series expansion?  Here is the problem: $$f(z)=(ze^{2-z}+(2-z)e^z)e^{(z-1)^{2}+(z-1)^{-2}}  $$ at z = 1  Thank you
42038
Mathematica tells me that $\sum\limits_{i=1}^n \frac1{2i-1}$ is equal to $\frac12 H_{n-1/2}+\log\,2$, where $H_n$ is a harmonic number.  Why is this true? Is there a general strategy for evaluating sums of the form $\sum\limits_{i=1}^n \frac1{ai+b}$?
103332
Can anyone help me with this?  What is the sum of this series:  $\frac{2}{\pi}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{2k-1}$  I got it after plugging $x=-1$ in a Fourier series   Thank you!
7410
For example how come $\zeta(2)=\sum_{n=1}^{\infty}n^{-2}=\frac{\pi^2}{6}$. It seems counter intuitive that you can add numbers in $\mathbb{Q}$ and get an irrational number.
105702
I know that $$\int^\infty_0\frac x{e^x-1}dx=\frac{\pi^2}{6}$$ For substituting $u=2$ into $$\zeta(u)\Gamma(u)=\int^\infty_0\frac{x^{u-1}}{e^x-1}dx$$  However, I suspect that there is an easier proof, maybe by the use of complex analysis. I haven't learnt the  yet. All I know is the above formula and $\zeta(2)=\frac{\pi^2}{6}$. Can we can use the above integral to find out some of the $\zeta$'s value?
39847
How to prove  $$\zeta(2)-\log(x)\log(1-x)=\operatorname{Li}_2(x)+\operatorname{Li}_2(1-x)$$ I havent started, any hints?
61068
Say a function's domain is $\{-\pi/2, \pi/2\}$. How should I interpret this interval?  It starts from where? To where? In what direction?  And if it starts from $3\pi/2$, would the next one be $-5\pi/3$.(because it starts from negative, $-\pi/2$).
3313
How do I evaluate this function for given $s$?  $$\zeta(s) = \sum_{n=1}^\infty \frac1{n^s} = \frac{1}{1^s} + \frac{1}{2^s} + \frac{1}{3^s} + \cdots$$
54040
I've been teaching my 10yo son some (for me, anyway) pretty advanced mathematics recently and he stumped me with a question. The background is this.  In the domain of natural numbers, addition and multiplication always generate natural numbers, staying in the same domain.  However subtraction of a large number from a smaller one needs to "escape" into the domain of integers, and division may result in escape to the real domain (like 3 / 5 -&gt; 0.6).  It was a simple step from there to taking the square root of a negative number, hence requiring the escape into the complex domain, such as 4+7i.  He quite easily picked up that each of these domains was a superset of another, natural -&gt; integer -&gt; real -&gt; complex.  However, he then asked if an operation on a complex number would require yet another escape, a question I had to investigate. Now, it turns out that the square root of a complex number is simply another complex number along the lines of mathematical distribution: (a+bi)2 -> a2 + 2abi - b2, from memory.  But I'm wondering if there are other mathematical operations performed on complex numbers (or any of its subset domains) that can't be represented within the complex domain.  Apologies if I've used the wrong terms, it's been about 30 years since I did University level math.
59255
One of the joys of high-school mathematics is summing a complicated series to get a “closed-form” expression. And of course many of us have tried summing the harmonic series $H_n =\sum \limits_{k \leq n} \frac{1}{k}$, and failed. But should we necessarily fail?      More precisely, is it known that $H_n$ cannot be written in terms of the elementary functions, say, the rational functions, $\exp(x)$ and $\ln x$? If so, how is such a theorem proved?    Note. When I started writing the question, I was going to ask if it is known that the harmonic function cannot be represented simply as a rational function? But this is easy to see, since $H_n$ grows like $\ln n+O(1)$, whereas no rational function grows logarithmically.  Added note: This  asks a similar question for “elementary integration”. I guess I am asking if there is an analogous theory of “elementary summation”.
100559
I must find the residues of $z^2\sin(\frac{1}{z})$ at $z = 0$.  Since $z = 0$ seems to be an Essential Singularity, i'm not sure how I can continue to find the residue of the function. Usually I am able to apply the Taylor Series and then find the $z^{-1}$ coefficient, but in this case I do not get a $z^{-1}$ term.
21083
The following appears as the second-to-last problem of Stewart's Complex Analysis:     Describe the Riemann surface of the function $y=\sqrt{z+z^2+z^4+\cdots +z^{2^n}+\cdots}$.   This problem intimidated me when I first saw it as an undergrad, as the series under the root isn't the expansion of any function I know (then or now). That renders the Riemann surface more subtle than usual, since any info about poles and zeroes must be found from this series alone. For instance, it has a zero at $z=0$ and therefore a branch point. But there should be other zeroes — indeed, infinitely many zeroes— since the number associated with any truncation of the series grows as $2^N$!. This suggests that the Riemann surface of this function must be exotic.  I'll give a sketch of an argument below, taking the pedestrian approach of hunting poles and zeroes of the function. However, I'd be glad to see an elegant and advanced perspective of the problem, particularly if it uses some ideas which I wouldn't have encountered as an undergraduate. The more food for thought, the better!  EDIT: As noted by O.L.'s answer below, the series above is the canonical example of a lacunary function with the unit circle as natural boundary. What that leaves open is the nature of the resulting Riemann surface, which would seem to hinge upon the quantity of the function's zeros within the unit circle. David Speyer has given an argument in his answer that this number is at least 10. Is there a more definitive result on the zeroes of this lacuuary function?