Lets define the following function series (Riemann Zeta Function):  $$\varsigma (z)=\sum _{n=1}^\infty n^{-z} $$  where the power is defined by the primary branch of the Logarithm.  A) prove that the series converges in : $ D= \left\{ z:\operatorname{Re}(z) &gt; 1 \right\}$ and also converges absolutely and uniformly in $D_\delta = \left\{z:\operatorname{Re}(z)&gt;1+ \delta \right\}  $  B) find $\varsigma '(z) $.  Do guys have any hints? I thought about calculating the radius of convergence, but how do I do it in this case? and how do I prvoe that it converges with this minus power? also I didnt get the information about the power being defined that way, im not sure also that I translated this accurately from hebrew.
121604
I'm aware there are been several similar questions, and some great answers with hints on how to prove this, for example  and .  But I've never really seen a detailed proof of the absolute convergence of the Riemann zeta function in the half-plane $\Re(s)&gt; 1$.  I hope there's interest on a detailed proof of this for reference. Here is my attempt. Please, fill in any detail that might be missing, or point out any mistake!  On $\Re(s)=\sigma&gt; 1$, we have  $$\sum_{n=1}^{\infty}\bigg|\frac{1}{n^s}\bigg|=\sum_{n=1}^{\infty}\frac{1}{n^\sigma}$$  and  $$\frac{1}{n^\sigma}\leq\frac{1}{n^{1+\epsilon}}$$  for any $\epsilon &gt;0$, so by the direct comparison test, if   $$\sum_{n=1}^{\infty}\frac{1}{n^{1+\epsilon}}$$  converges absolutely, then so do $\sum_{n=1}^{\infty}|1/n^s|$ and $\zeta(s)$.  By the integral test, the convergence of $\sum_{n=1}^{\infty}1/n^{1+\epsilon}$ is equivalente to the finitude of the integral  $$\int_1^\infty \frac{1}{n^{1+\epsilon}}=\bigg[-\frac{1}{\epsilon x^\epsilon}\bigg]=\frac{1}{\epsilon}$$  which holds for any $\epsilon &lt; \infty$.  Any correction or improvement is welcomed! Thanks in advance.
19946
If I have $\sum_{n=2}^{\infty} \frac {1}{n\log n}$ and want to prove that it diverges, can I use following?  $$\frac {1}{n\log n} \lt \frac {1}{n}$$   $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of $\frac 1 n$ equals to zero so the comparasion I think isn't useful. Or can I say it diverges because $\sum_{n=1}^\infty \frac 1{\log n}$ also diverges?   And If I have $\sum_{n=2}^\infty \frac {(-1)^n}{(n)}$ I can use the Leibniz rule. $\sum_{n=1}^\infty \frac 1 n$ diverges, but the limit of it equals to zero, so I am confused if this series diverges/converges again. I understand that $\sum_{n=1}^\infty \frac 1 n$ has an infinite sum, but what to do in this case?
13596
I can not find the function from which I have to start to calculate this power series.  $$\sum_{n=1}^\infty nx^n$$ Any tips?. Thanks.
69137
Does $\sum\frac{(-1)^{\lfloor \sqrt n\rfloor}}{n}$ converge ?  I know there is a technical method that uses an adapted Abel's rule  $(*)$ ; I would like to know if you see another method (which is understable with basic knowledge) to prove it converges. Thank you.  Remark: Actually, it can be proved that for all $z \in \mathbb C$ such that $|z| = 1$, $\sum\frac{(-1)^{\lfloor \sqrt n\rfloor}}{ n}z^n$ converges.  $(*)$ Consider two series $\sum a_n$ and $\sum b_n$. Define $B_m = \sum_{n=0}^mb_n$   If the sequence $(a_n)$ tends to $0$ and the sequence $(B_n)$ is bonded, and if the series $\sum (a_{n+1} - a_n)$ converges absolutely, then $\sum a_nb_n$ converges.  You have to adapt this theorem to use it here.
45695
Could anyone please give me some hints ?   Let $\sum a_n$ be a convergent serie of real number. Prove or disprove that $\sum a_n \sin n$ and $\sum n^{\frac{1}{n}} a_n$ are also convergent.
23507
I just came across the problem of what happens in the boundary cases of .  More concisely, for what $\theta$ will the following sum converge? $$\sum_{n=1}^{\infty} \frac{e^{ni\theta}}{n}$$ It is well known that this sum will diverge for $\theta=0$ and will converge for $\theta=\pi$. In fact, for any odd multiple of an even root of unity, since we have symmetry, for each line through the origin of the complex plane, we can use the alternating series theorem to show that they converge.  Another idea I had was to split  the sum into imaginary and real parts, and if we can prove each of these converges, then we are all set. So now we must prove that both of the following converge $$\sum_{n=1}^{\infty}\frac{\cos(n\theta)}{n}\space, \space \sum_{n=1}^{\infty}\frac{\sin(n\theta)}{n}$$ Now these can be shown to converge easily by Dirichlet's test for any rational multiple of pi, but what about irrational multiples of pi? I know it has been shown to converge with $\theta=1$, but I don't see how to generalize this.
25401
For any $x\in\mathbb{R}$, the series $$ \sum_{n\geq 1}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) $$ is trivially absolutely convergent. It defines a function $f(x)$ and I would like to show that $f(x)$ is unbounded over $\mathbb{R}$. Here there are my thoughts/attempts:   $$(\mathcal{L} f)(s) = \sum_{n\geq 1}\frac{1}{1+n^2 s^2} = \frac{-s+\pi\coth\frac{\pi}{s}}{2s}=\sum_{m\geq 1}\frac{(-1)^{m+1}\,\zeta(2m)}{s^{2m}}$$ is a function with no secrets. It behaves like $\frac{\pi}{2s}$ in a right neighbourhood of the origin, like $\frac{\pi^2}{6s^2}$ in a left neighbourhood of $+\infty$. The origin is an essential singularity and there are simple poles at each $s$ of the form $\pm\frac{i}{m}$ with $m\in\mathbb{N}^+$. These facts do not seem to rule out the possibility that $f$ is bounded; For any $N\in\mathbb{N}^+$ there clearly is some $x\ll e^N$ such that $\sin(x),\sin\left(\frac{x}{2}\right),\ldots,\sin\left(\frac{x}{N}\right)$ are all positive and large enough, making a partial sum of $ \sum_{n\geq 1}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) $ pretty close to $C\log N$. On the other hand I do not see an effective way for controlling  $ \sum_{n&gt;N}\tfrac{1}{n}\,\sin\left(\tfrac{x}{n}\right) $ - maybe by summation by parts, by exploiting the bounded-ness of the sine integral function? Some probabilistic argument might be effective. For any $n\geq 3$ we may define $E_n$ as the set of $x\in\mathbb{R}^+$ such that $\sin\left(\frac{x}{n}\right)\geq \frac{1}{\log n}$. The density of any $E_n$ in $\mathbb{R}^+$ is close to $\frac{1}{2}$, so by a Borel-Cantellish argument it looks reasonable that the set of points such that $|f(x)|\geq \frac{\log x}{100}$ is unbounded, but how to make it really rigorous? To compute $\lim_{x\to x_0}f(x)$ through convolutions with approximate identities seems doable but not really appealing.
84143
Consider the series $ \sum_{n=1}^\infty x^2+ n/n^2$  . Pick out the true statements: (a) The series converges for all real values of $x$. (b) The series converges uniformly on $\mathbb{R}$. (c) The series does not converge absolutely for any real value of $x$.        stuck on this problem totally. please help me to solve this problem.thanks .
24878
I am given a series: $$\sum\limits_{n=1}^\infty \sin\left(\pi \sqrt{n^2+\alpha}\right)$$ And in the description of the problem it is said that I must show by Leibniz test that it is convergent. How can I even get the $(-1)^n$ symbol out of the starting series?
71873
A while ago, there was a great hype about the “identity” $$\sum_{n=1}^{\infty} n = -\frac{1}{12}.$$  Apart from some series manipulations where the validity seems to be at least questionable, the derivation of this always goes through the zeta function:  Where the series converges, the zeta function is defined by $$\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s}$$ and outside that range by analytic continuation. And it turns out that inserting $s=-1$ formally results in $$\zeta(-1) = -\frac{1}{12} = \sum_{n=1}^\infty n$$  However looking at the series in isolation, there is no indication that the zeta function should be chosen.  An obvious way to get an analytic function that at one point gives the sum of all natural numbers is $$f(x) = \sum_{n=1}^{\infty} nx^n$$ at $x=1$, however (not surprisingly) that function diverges at $1$.  Therefore my question:     Is it possible to get another finite value for the series by analytic continuation of another series?   Concretely, do there exist continuous functions $f_1, f_2, f_3, \ldots$ such that   On some non-empty open subset $S$ of $\mathbb C$, $f(x)=\sum_{n=1}^\infty f_n(x)$ converges to an analytic function. At some point $x_0$, $f_n(x_0) = n$ for all positive integers $n$. The analytic continuation of $f$ is well defined and finite at $x_0$. $f(x_0) \ne -1/12$   What if we demand the functions $f_n$ to be analytic rather than just continuous?
55167
let $m$ is give a positive integers,   Determine for which values of $a$,the series $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}$$ converges     where $[x]$ is the largest integer not greater than $x$.  and  I have see this not hard problem :show that $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}$$ converges, solution: note     $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}=\sum_{n=1}^{\infty}(-1)^n\left(\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}\right)$$   and we have   let $$a_{n}=\dfrac{1}{n^2}+\dfrac{1}{n^2+1}+\cdots+\dfrac{1}{n^2+2n}$$   $$0&lt;a_{n}&lt;\dfrac{2n+1}{n^2}&lt;\dfrac{3}{n}\longrightarrow 0$$   and   $$a_{n}-a_{n+1}=\dfrac{1}{n^2}-\dfrac{1}{(n+1)^2}+\dfrac{1}{n^2+1}-\dfrac{1}{(n+1)^2+1}+\cdots+\dfrac{1}{n^2+2n}-\dfrac{1}{(n+1)^2+2n}-\left(\dfrac{1}{(n+1)^2+2n+1}+\dfrac{1}{(n+1)^2+2(n+1)}\right)$$   so   $$a_{n}-a_{n+1}=(2n+1)\left(\dfrac{1}{n^2(n+1)^2}+\cdots+\dfrac{1}{(n^2+2n)(n^2+4n+1)}\right)-\dfrac{1}{n^2+4n+2}-\dfrac{1}{n^2+4n+3}$$   so   $$a_{n}-a_{n+1}&gt;\dfrac{(2n+1)^2}{(n^2+2n)(n^2+4n+1)}-\dfrac{2}{n^2+4n+1}=\dfrac{2n^2+1}{(n^2+2n)(n^2+4n+1)}&gt;0$$   so   $$\sum_{n=1}^{\infty}(-1)^na_{n}=\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt{n}]}}{n}$$ converges   But for my problem,I can't   My try:  $$\sum_{n=1}^{\infty}\dfrac{(-1)^{[\sqrt[m]{n}]}}{n^a}=\sum_{n=1}^{\infty}(-1)^n\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}$$ let $$b_{n}=\sum_{k=n^m}^{(n+1)^m-1}\dfrac{1}{k^a}$$ then how prove  $$b_{n}\longrightarrow 0$$and $$b_{n}&gt;b_{n+1}$$
34642
How to evaluate the series: $$  2 \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n+1}\left( 1 + \frac12 + \cdots + \frac 1n\right) $$  According to Mathematica, this converges to $ (\log 2)^2 $.
73045
I have a problem with solving the convergence of $\sum_{n=1}^{\infty }\frac{(-1)^{[\sqrt{n}]}}{n}$ where $[a]$ means floor function of $a$. I've already made the first thep and rewrote it with the $\sum_{n=1}^{\infty }(-1)^{n}\sum_{k=n^2}^{(n+1)^2-1}\frac{1}{k}$, which converges if and only if the previous one converges. Now I want to use Leibniz criterion, but I cannot prove that the second sum in the previous expression is monotonic. Does anyone know how to do it? (I cannot use Riemann integral, we haven't defined it yet)
68008
For $x\in (-\pi, \pi) $, I want to know whether the following series following series are uniformly convergent on  $(-\pi,\pi) $: 1. $\sum _{n=1}^{\infty}{(\frac {x}{n})^n} $ 2. $\sum_{n=1}^{\infty}{\frac {1}{((x+\pi)n)^2}}$. I think series in 2 has to be uniformly convergent for, $|\frac{1}{((x+\pi)n)^2}|$ $\leq $ $\frac {1}{n^2} $ and as $\sum {\frac {1}{n^2}} $ is convergent, by Wierstrass M-test the series in (2) is uniformly convergent. But I don't how to proceed for the series in (1). Any help please!  Also tell if my answer to the (2)nd is correct.
9865
I think the radius of convergence for  $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)x^n$, $x\in \mathbb R$ is:   $r^{-1}$=$\lim_{n\to \infty}$$|\frac{a_{n+1}}{a_n}$|=1   so we get that $r$=1.  But how can I show it formally?  After that, I have to show that $\displaystyle\sum_{n=1}^{\infty}\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)x^n$ is uniformly convergent in the interval $[-r,r]$. Can I maybe use Weierstrass majoranttest? What can I use as majorant serie?
41043
I'm confused how this series :$$\displaystyle \sum_{n=1}^\infty\frac1{n\sin n}$$ is a convergent series as wolfram show  and it's not harmonic series in the same time .  My question here is:  Why this :$$\displaystyle \sum_{n=1}^\infty\frac1{n\sin n}$$ is not harmonic series and is convergent ?
61887
I know that the series $\sum_{k=1}^\infty \frac{\sin k}{k}$ converges (to $\frac{\pi - 1}{2}$), though by crazy stuff with Dirichlet Kernels or by reverse-engineering $\frac{\pi - x}{2} = \sum \frac{\sin (kx)}{k}$ using Fourier Series.  My question instead: Is $\sum_{k=1}^\infty \frac{\sin k}{k}$ absolutely convergent?  My gut instinct is no: It behaves not-much-better than $\sum \frac1k$, but that's certainly not a rigorous proof.