5
votes
2answers
107 views

A series $\sum_{j=1}^{\infty}\sum_{i=1}^{\infty}\frac{(i-1)! (j-1)!}{(i+j)!}H_{i+j}$ and $\zeta(3)$

We have $$ \sum_{j=1}^{\infty}\sum_{i=1}^{\infty} \displaystyle \frac{(i-1)! (j-1)!}{(i+j)!} H_{i+j} =\displaystyle 3 \: \zeta(3) $$ where $\displaystyle H_{n}:=\sum_{1}^{n} \frac{1}{k}$ are ...
4
votes
2answers
91 views

What is $\zeta(n)$ as $n$ tends to $\infty$? How fast it goes to the limit?

What is $\zeta(n)$ as $n\to\infty$? How fast it goes to the limit?
5
votes
2answers
120 views

From the series $\sum_{n=1}^{+ \infty} \left(H_{n}-\ln n-\gamma -\frac{1}{2n}\right)$ to $\zeta(\frac{1}{2}+it)$

Here is a pretty series $$ \displaystyle \sum_{n=1}^{+ \infty} \left(H_{n}-\ln n-\gamma -\frac{1}{2n}\right)=\frac{1}{2} \left(1-\ln (2\pi)+\gamma\right) \quad (*) $$ where $H_{n}:=\sum_{1}^{n} ...
3
votes
0answers
74 views

An interesting identity involving powers of $\pi$ and alternating zeta series

It happens that, for any $m\geq 1$, $$\sum_{n=0}^{+\infty}\frac{(-1)^n}{(2n+1)^{2m+1}}=\frac{E_{2m}}{2\cdot(2m)!}\left(\frac{\pi}{2}\right)^{2m+1}\tag{1}$$ where $E_{2m}$ is an integer number. My ...
0
votes
1answer
58 views

Question about the convergence of an infinite series in all of $\mathbb{C}$.

Does the following infinite series: $$G(s):=\displaystyle \sum^{\infty}_{n=1} \frac{(-1)^{n+1}}{n^s + \frac{1}{n^s}}$$ converge for all $s \in \mathbb{C}$ (especially in the critical strip; since ...
1
vote
1answer
62 views

Investigating the convergence of a series using the comparison limit test, Part II [duplicate]

I posted this question earlier, but as I don't know if a comment reply or edit will refresh this so people actually see, I'm going to post it again in hopes that someone knows what's going on. Here's ...
0
votes
0answers
31 views

Logarithmic derivative of Riemann zeta, is this derivation correct?

Let matrix $T_2$ be defined below as the Dirichlet inverse of the Euler totient function as a function of the Greatest Common Divisor (GCD) of row index $n$ and column index $k$; $$T_2(n,k) = ...
0
votes
0answers
66 views

Evaluating $\int_2^\infty \zeta(x) - 1 \,\, \mathrm{d}x$

While looking at a table of values for the zeta function, the fact that they approach $1$ made me wonder what the improper integral of the fractional part of the zeta function would be. I've found ...
5
votes
0answers
101 views

Asymptotic expansion of $\zeta(s)$

It is well known that $$ \zeta(s) = \sum_{n = 1}^\infty \frac{1}{n^s} = \prod_{p \text{ prime}} \frac{1}{1 - p^{-s}}, \quad \Re[s] > 1, \tag{1}$$ but, if $p \leq N$ denotes the primes less than or ...
1
vote
2answers
80 views

Find the sum $\sum_{n = 1}^{\infty}(-1)^{n + 1}\log(1 + (1/n))$

I started as follows $$\begin{aligned}S &= \sum_{n = 1}^{\infty}(-1)^{n + 1}\log\left(1 + \frac{1}{n}\right)\\ &= \sum_{n = 1}^{\infty}(-1)^{n + 1}\sum_{k = 1}^{\infty}(-1)^{k + 1}\frac{1}{k ...
5
votes
2answers
112 views

$\prod_{i=1}^{\infty}{1+(\frac{k}{i})^3}$ for integer k

Can anyone compute $$\prod_{i=1}^{\infty}{1+(\frac{k}{i})^3}$$ for integer k? Can it be done in closed form, using only elementary functions, without the use of the Gamma function? For k=1, the closed ...
1
vote
2answers
133 views

Information about Riemann Zeta function

I have general question on Riemann Zeta function. How can I improve knowledge on Riemann Zeta Function theory up to research? For example , what are the best books on Zeta ...
4
votes
3answers
606 views

$1 + 1 + 1 +\cdots = -\frac{1}{2}$ [closed]

The formal series $$ \sum_{n=1}^\infty 1 = 1+1+1+\dots=-\frac{1}{2} $$ comes from the analytical continuation of the Riemann zeta function $\zeta (s)$ at $s=0$ and it is used in String Theory. I am ...
2
votes
0answers
104 views

A connection between a sequence from the Collatz conjecture and a sequence of densities from $\zeta(k)-1$?

Just for grins, I created lists of first-entries of finite sequences of rank $r$ for the Syracuse problem (Collatz conjecture using only odd numbers) and found these sequences on OEIS. My sequences, ...
2
votes
0answers
91 views

Series involving the Riemann zeta function

Consider the series: $$\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{n(2n+1)}$$ We can easily prove that it's a convergent series. My question, is there a way to express this series in terms of zeta ...
1
vote
1answer
135 views

Closed form for $\sum_{k=1}^{\infty} \zeta(2k)-\zeta(2k+1)$

From WolframAlpha it seems that $$ \frac{1}{2}=\sum_{k=1}^{\infty} \zeta(2k)-\zeta(2k+1) $$ Could someone provide a proof for this? Thanks.
1
vote
3answers
251 views

Does $\zeta(-1)=-1/12$ or $\zeta(-1) \to -1/12$? [duplicate]

I saw NumberPhile channel on Youtube, and they proved $1+2+3+\cdots=-1/12$. Also, I read This. So, which one is correct $$\zeta(-1)=-1/12\\ \text{or} \\\zeta(-1) \to -1/12$$ Equivalent to: ...
35
votes
3answers
1k views

Does $\sum _{n=1}^{\infty } \frac{\sin(\text{ln}(n))}{n}$ converge?

Does $\sum _{n=1}^{\infty } \dfrac{\sin(\text{ln}(n))}{n}$ converge? My hypothesis is that it doesn't , but I don't know how to prove it. $ζ(1+i)$ does not converge but it doesn't solve problem here. ...
0
votes
2answers
135 views

Value of Riemann zeta function at $-1$

This claim is false $\sum_{n=1}^{\infty}n=\sum_{n=1}^{\infty}n^{-(-1)}= \zeta(-1)=-1/12$. The error is that we should $\sum_{n=1}^{\infty}n=\sum_{n=1}^{\infty}(1/n ^1)^{-1}=(0)^{-1}$. Am I correct? ...
8
votes
4answers
355 views

Proving that $\frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}}$

After numerical analysis it seems that $$ \frac{\pi^{3}}{32}=1-\sum_{k=1}^{\infty}\frac{2k(2k+1)\zeta(2k+2)}{4^{2k+2}} $$ Could someone prove the validity of such identity?
12
votes
3answers
281 views

Proving that $\left(\frac{\pi}{2}\right)^{2}=1+\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k-1}}$.

Wolfram$\alpha$ says that we have the following identity $$ \left(\frac{\pi}{2}\right)^{2}=1+2\sum_{k=1}^{\infty}\frac{(2k-1)\zeta(2k)}{2^{2k}} $$ but, how does one prove such identity?
10
votes
2answers
201 views

Evaluating $\sum_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}$

I want to find $$\sum_{n=1}^\infty\frac{\zeta(3n)}{2^{3n}}$$ I let $f(z)=\sum_{n=1}^\infty\frac{1}{2^{3n}}z^{3n}$ and now $$\sum_{n=1}^\infty ...
10
votes
1answer
208 views

Prove $\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$

How can I prove that $$\sum_{n=1}^\infty\frac{H_{2n+1}}{n^2}=\frac{11}{4}\zeta(3)+\zeta(2)+4\log(2)-4$$ I think this post can help me, but I'm not sure. Mathematica can provide a closed form for ...
1
vote
1answer
56 views

Series divergence - how precise should the answer be

Morning. I've written down some of my reasoning and arguments as to why the series diverges, however I am not certain I can safely conclude it diverges to $\infty$. Would you give it a look, please? ...
3
votes
1answer
83 views

Sum of reciprocal of zeroes of zeta function

The sum of reciprocal of zeroes of riemann zeta function converges conditionally that if they are paired as $\rho $ and $1-\rho$ My question is if the sum still converges if they are paired as $\rho$ ...
30
votes
3answers
628 views

proving that $\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$

Prove that $$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$ ($H_n=\sum_{k=1}^{n}\frac{1}{k}$)
6
votes
3answers
225 views

Question about Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$

For a freshman calculus project, I used Euler's approach to find $\sum_{n=1}^\infty\frac1{n^2}=\frac{\pi^2}6$, and noted from Wikipedia's explanation that the infinite product representation of ...
7
votes
1answer
92 views

closed form of $\frac{(-\pi)^m\zeta(m,\frac{x}{\pi})+\pi^m\zeta(m,1-\frac{x}{\pi})}{(-1)^{m}\pi^{2m}}$

I know that following equality is true. $$\sum_{n=-\infty}^{\infty}\frac{1}{(x+n\pi)^m}=\frac{(-\pi)^m\zeta(m,\frac{x}{\pi})+\pi^m\zeta(m,1-\frac{x}{\pi})}{(-1)^{m}\pi^{2m}}$$ But can we find the ...
3
votes
2answers
141 views

Proof of my conjecture on closed form of $\int _{0}^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}$

Let $a$, $b\in \Bbb R^+$ and $m \in \Bbb N$ then My conjectural closed form is $$\int _{0}^{\infty}\frac{x^{a-1}e^{-mbx}}{1-e^{-bx}}\,{\rm d}x = ...
0
votes
0answers
139 views

Is there a elementary way to prove $\zeta(2)=\frac{\pi^2}{6}$

The proof in the Wikipedia is still much complicated, can any one provide a really simple way to prove this.
1
vote
1answer
55 views

How does $\zeta(i\pi)$ converge?

$$\zeta(i\pi) = \sum_{r=1}^{\infty}r^{-i\pi} = \sum_{r=1}^{\infty}e^{-i\pi \ln(r)} = \sum_{r=1}^{\infty}\operatorname{cis}(-\pi\ln(r))$$ Did I mess up somewhere in the steps above? I can't see how ...
10
votes
0answers
341 views

I can Euler-sum $\sqrt{-\ln(1)}-\sqrt{-\ln(2)}+\sqrt{-\ln(3)}-\cdots$. But how can I do $\sqrt{-\ln(1)}+\sqrt{-\ln(2)}+\sqrt{-\ln(3))}+\cdots$?

This is also related to an older thread in MSE ("what is the half derivative of zeta at zero?") . One of the possible steps in the problem of that thread was to evaluate the series ...
23
votes
6answers
956 views

Evaluating $‎\sum_{n=2}^{\infty}\frac{\zeta(n)}{k^n}$

‎If $f\left(z \right)=\sum_{n=2}^{\infty}a_{n}z^n$ and $\sum_{n=2}^{\infty}|a_n|$ converges then‎, $$\sum_{n=1}^{\infty}f\left(\frac{1}{n}\right)=\sum_{n=2}^{\infty}a_n\zeta\left(n\right)‎.$$ ‎Since ...
2
votes
1answer
71 views

Show in between steps in this Riemann zeta function equivalence/reduciton

In the answer chosen by the OP in this question I had trouble understanding the steps taken to get the equivalences/reduce the zeta function into another one. Can somebody show me the steps to go from ...
10
votes
2answers
398 views

Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$

I found the following formula $$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$ and it is cited that Euler proved the ...
5
votes
2answers
219 views

Is there any value of zeta that is an integer?

Is there any value which we can substitute for $s$ in $\zeta (s)$ such that $$\sum_{n=1}^{\infty }n^{-s}\in \mathbb{Z}$$
3
votes
1answer
211 views

Proving that $\sum (-1)^{n+1} n^{-z}$ defines an analytic function in $Re z>0$

I want to show that the series $\sum_{n=1}^\infty (-1)^{n+1} n^{-z}$ converges to an analytic function for $\Re z>0$. For $\Re z>1$ the terms are dominated by $n^{-x}$ so that we have absolute ...
12
votes
1answer
330 views

Proof of $\sum_{n=1}^\infty \frac{1}{n^4 \binom{2n}{n}}=\frac{17\pi^4}{3240}$

Recently, I was able to prove that $$\sum_{n=1}^\infty \frac{1}{n \binom{2n}{n}}= \frac{\pi}{3\sqrt{3}}$$ $$\sum_{n=1}^\infty \frac{1}{n^2 \binom{2n}{n}}= \frac{\pi^2}{18}$$ But does anybody know ...
0
votes
4answers
114 views

Let ${P_n}$ be the sequence of all consecutive prime numbers. Is $\sum_{n\geq 1} \frac{1}{p_n}$ convergent? [duplicate]

Let ${P_n}$ be the sequence of all consecutive prime numbers. Is $\sum_{n\geq 1}\frac{1}{p_n}$ convergent?
6
votes
3answers
250 views

What is the half-derivative of zeta at $s=0$ (and how to compute it)?

[Update 3:] I gave a new partial answer following the ansatz in question Q3. I leave the other parts of the question untouched, they are also partially answered in specialized other questions in MSE. ...
12
votes
5answers
338 views

Alternating sum of multiple zetas equals always 1?

This is more in the category of "recreational math"... I was playing with multiple zetas, in the notation of $\zeta(k),\zeta(k,k),\zeta(k,k,k),\ldots$ as given in wikipedia. Looking at the alternating ...
13
votes
1answer
282 views

A Tough Series $\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$

I have done series with $\zeta(2k)$ and $\zeta(k)$, but I have no idea with this one: $$\sum_{k=1}^\infty \frac{\zeta(2k+1)-1}{k+1}=-\gamma+\log(2)$$ $\gamma$ is the Euler–Mascheroni Constant. This ...
8
votes
4answers
400 views

Sum : $\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^3}$

Prove that : $$\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)^3}=\frac{\pi^3}{32}.$$ I think this is known (see here), I appreciate any hint or link for the solution (or the full solution).
1
vote
0answers
37 views

The sum of the reciprocals of fourth powers [duplicate]

This problem is an extension of the well known basel problem and involves finding the sum of 1 + 1/16 + 1/81 ... = 1/1^4 + 1/2^4 + 1/3^4 ... 1/n^4 where n tends to infinity Euler managed to prove ...
3
votes
1answer
175 views

Discontinuities of $\sum \frac{x^{\rho}}{\rho}$

H. Edwards in his book on the zeta function says that $\sum\frac{x^{\rho}}{\rho}$ converges conditionally "even when $\rho ,1-\rho$ are paired." I tried calculating some terms (n = 500 or so) and ...
12
votes
2answers
357 views

Can the Basel problem be solved by Leibniz today?

It is well known that Leibniz derived the series $$\begin{align} \frac{\pi}{4}&=\sum_{i=0}^\infty \frac{(-1)^i}{2i+1},\tag{1} \end{align}$$ but apparently he did not prove that $$\begin{align} ...
14
votes
3answers
473 views

A series involves harmonic number

How do we get a closed form for $$\sum_{n=1}^\infty\frac{H_n}{(2n+1)^2}$$
4
votes
1answer
156 views

About Euler's formula for Apery number

Euler's formula. $$\zeta(3)=\frac{\pi^2}{7}\left(1-4\sum_{m\ge 1}\frac{\zeta(2m)}{(2m+1)(2m+2)2^{2m}}\right)$$ I saw this formula in Wikipedia a few months ago. I have searched about Euler's ...
7
votes
2answers
342 views

A tough series: $\sum_{k=1}^{\infty}\frac{\zeta(2k)}{(k+1)(2k+1)}$, need help

I was doing a integral which ends up with a tough series part: $$\sum_{k=1}^{\infty}\frac{\zeta(2k)}{(k+1)(2k+1)}$$ Mathematica says $$\frac12$$ Which agrees with the anwer...Anyone know how to ...
3
votes
2answers
148 views

Another improper integral

Show that : $$\int_0^1\frac{(\sin ^{-1}x)^2}{x}\text{d}x=\frac{\pi ^2\ln 2}{4}-\frac78\zeta(3)$$ This integral is in "irresistible integrals" on page 122. I can't prove this one.