All Questions
Tagged with harmonic-numbers generating-functions
33 questions
7
votes
1
answer
462
views
Ramanujan-Type Harmonic Series $\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)H_{r-1/2}=-\frac{8\ln2}{9\pi}$
We will be considering series such as this :
$$\sum_{r=0}^{\infty}\binom{2r}{r}^3\left(\frac{1/6+r}{2^{8r}}\right)H_{r-1/2}=-\frac{8\ln2}{9\pi}$$
Consider $K(k)$ and $E(k)$ to be the Complete ...
- 3,710
8
votes
1
answer
317
views
Closed form of $\sum _{n=2}^{\infty }\zeta (n)^2 z^n$
Recently, interesting problems related to sums over the zeta function reappeared, see e.g. [1] or [2].
In approaching these problems it is useful to study the generating sums. The generating sum of ...
2
votes
0
answers
88
views
Can the generating function $\sum_{n = 1}^{\infty} H_n^k z^n$ for $k \in \mathbb{N}$ generally be determined by special functions?
I have been playing around a lot with Euler sums lately, and in an attempt to derive results not explicitly mentioned in the papers I've read, I wanted to get the OGF of
$$\sum_{n = 1}^{\infty} H_n^4 ...
- 1,459
7
votes
4
answers
652
views
Coefficient extraction,show: $[z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha})$
I want to show:
\begin{equation*}
[z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th ...
- 1,070
1
vote
1
answer
130
views
An identity involving the central binomial coefficient and the harmonic numbers
While doing some computational experiments, I found the following (conjectured) identity:
$$2^{2n-1}\sum_{k=0}^{\infty}(-1)^{k}\frac{a_{n,k}}{2^k}=\binom{2n}{n}$$
where $a_{k}$ is defined as
$$a_{n,k+...
3
votes
2
answers
339
views
How to prove this identity on exponential generating function of harmonic numbers
I came across the following problem, let $N![z^N]A(z)$ denote the coefficient of an exponential generating function (EGF) $A(z)$. The EGF is similar to an ordinary generating function (OGF) $A'(z)$ ...
- 655
0
votes
3
answers
193
views
Mustering $\sum_{k=1}^{\infty} \frac{(-1)^kH_{2k}}{k^2}$ with complex series
I have a series
$$\sum_{k=1}^{\infty} \frac{(-1)^kH_{2k}}{k^2}$$
And I attempted to use the generating function
$$\sum_{k=1}^{\infty} \frac{x^kH_k}{k^2} = \mathrm{Li}_2(1-x)\ln(1-x) + \frac{1}{2}\...
- 1,259
3
votes
2
answers
194
views
Sum $\sum_{k=1}^{\infty} \frac{(-1)^{k-1}H_{2k}}{k}$
I am particularly interested in solving
$$\sum_{k=1}^{\infty} \frac{(-1)^{k-1}H_{2k}}{k}$$
Where $$H_n = \sum_{k=1}^n \frac{1}{k}$$
I can’t seem to crack it. As much as I’d love to use the ...
- 1,259
3
votes
1
answer
148
views
How to prove $\sum _{k=1}^{\infty } (-1)^k H_{\frac{2 k}{3}} = -\frac{\pi }{2 \sqrt{3}}+\frac{3 \pi }{8}-\frac{3}{4} \log (2)$?
I stumbled on this problem in the wake of the discussion https://math.stackexchange.com/a/3553902/198592
Can you make sense of the equation in the question involving the harmonic number with a ...
5
votes
3
answers
321
views
Is the closed form for $\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}$ known in the literature?
I managed to find
$$\sum_{k=1}^\infty\frac{\overline{H}_k}{k^m}=(1-2^{-m})\sum_{k=1}^\infty\frac{H_k}{k^m}-2^{-m}\sum_{k=1}^\infty\frac{H_k}{(k+1/2)^m}$$
$$=(1-2^{-m})\left[\left(1+\frac m2\...
- 26.6k
3
votes
1
answer
157
views
How to calculate the generating function of the unified harmonic sum $U(\sigma,n,p) = \sum_{k=1}^n \frac{\sigma^k}{k^p}$?
Euler sums are dealing usually with either the harmonic sum or the alternating harmonic sum. Most of the operations and result are, however, similar in many respects.
Hence I propose here to study ...
4
votes
2
answers
254
views
What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^3}x^n\ ?$
Recently, the generating function of order 2 for the alternating harmonic series was calculated (What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$).
I would ...
4
votes
4
answers
316
views
What's the generating function for $\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$
Is there closed form for
$$\sum_{n=1}^\infty\frac{\overline{H}_n}{n^2}x^n\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
My approach,
In this ...
- 26.6k
7
votes
1
answer
374
views
How to evaluate $\int_0^y\frac{\ln x\ln^2(1-x)}{x}dx$
How to find $$\int_0^y\frac{\ln x\ln^2(1-x)}{x}dx\ ?$$
I came across this integral while I was trying to find $\displaystyle \sum_{n=1}^\infty\frac{H_n}{n^3}x^n$ and here is my work,
Divide the ...
- 26.6k
5
votes
0
answers
207
views
What's the Maclaurin series of $\frac{\ln(1-x)\ln(1+x)}{1+x^2}$?
I am trying to find a closed-form for the Maclaurin series of $\frac{\ln(1-x)\ln(1+x)}{1+x^2}$. I am not sure if it's possible but here is what I did:
We know that:
$$\ln(1-x)\ln(1+x)=\sum_{n=1}^\...
- 26.6k
1
vote
2
answers
554
views
Find $[z^N]$ for $\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$
Find $[z^N]$ for $\frac{1}{1-z} \left(\ln \frac{1}{1-z}\right)^2$. Here generalized harmonic numbers should be used.
$$H^{(2)}_n = 1^2 + \frac{1}{2}^2 + \dots + \frac{1}{N}^2.$$
For now, I was able ...
- 1,705
9
votes
5
answers
2k
views
A group of important generating functions involving harmonic number.
How to prove the following identities:
$$\small{\sum_{n=1}^\infty\frac{H_{n}}{n^2}x^{n}=\operatorname{Li}_3(x)-\operatorname{Li}_3(1-x)+\ln(1-x)\operatorname{Li}_2(1-x)+\frac12\ln x\ln^2(1-x)+\zeta(3)...
- 26.6k
2
votes
1
answer
288
views
Generating function of series $\sum\limits_{n=1}^\infty H_n^{(m)}\binom{2n}{n}x^n$?
I wish to find the generating function of infinite series $$\sum\limits_{n=1}^\infty H_n^{(m)}\binom{2n}{n}x^n= ?,$$ where $H_n^{(m)}$ is generalized harmonic number defined by $$H_n^{(m)}:=\sum\...
- 537
2
votes
2
answers
143
views
Prove $\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$
Is there an elegant proof for this identity for all real $s \neq -1, -1/2$?
$$\sum_{n=0}^\infty (-1)^n \binom{s}{n} \frac{H_{n+1/2}+\log 4}{n+1} =\frac{2^{2s+1}}{(2s+1) (s+1) {2s \choose s}}$$
Where ...
- 32.2k
8
votes
3
answers
737
views
prove $\ln(1+x^2)\arctan x=-2\sum_{n=1}^\infty \frac{(-1)^n H_{2n}}{2n+1}x^{2n+1}$
I was able to prove the above identity using 1) Cauchy Product of Power series and 2) integration but the point of posting it here is to use it as a reference in our solutions.
other approaches ...
- 26.6k
0
votes
0
answers
93
views
first derivative of exponential generating function of harmonic numbers
I'm trying to prove that the the first derivative of the exponential generating function of the Harmonic numbers, $H_n$, is the exponential generating function of $H_{n+1}$, but I my solution seems ...
- 407
10
votes
2
answers
513
views
Infinite series with harmonic numbers related to elliptic integrals
It is known that the following functions are elliptic integrals
$$
\, _2F_1\left({a,1-a\atop 1};x\right),\quad a=\tfrac12,\tfrac13,\tfrac14,\tfrac16,\tag{1}
$$
$$
\, _2F_1\left({\tfrac{1}{3},\tfrac{2}{...
- 4,005
2
votes
0
answers
111
views
Generating function for harmonic number times log ($H_k log(k)$)
During my studies of generating functions for expressions composed of harmonic numbers and its asyptotic approximations I found a simple sum which seems to have been neglected so far
$$s171219a=\sum ...
7
votes
2
answers
763
views
Generating function of square harmonic numbers
During my study of generating functions, I was able to calculate the generating function of the sequence of harmonic numbers $H_n$:
$$\sum_{n=1}^\infty H_nx^n=\frac{\ln(1-x)}{x-1}$$
However, I also ...
- 40.3k
2
votes
4
answers
386
views
Harmonic number identity $\sum_{k=1}^{n}H_k=(n+1)(H_{n+1}-1)$
Prove using generating functions that for Harmonic numbers $H_n= \sum_{j=1}^{n}\frac{1}{j}$
$$\sum_{k=1}^{n}H_k=(n+1)(H_{n+1}-1) (*)$$
The generating function for harmonic numbers is $$\sum_{n \in \...
- 836
3
votes
1
answer
295
views
Riemann zeta function related to the complete exponential bell polynomial in terms of the generalized harmonic number.
Prove that
$$\zeta(s)=\frac 1 {s!(1-2^{1-s})} \sum_{n=1}^\infty \frac{B_s(0!H_n^{(1)},1!H_n^{(2)},\dots,(s-1)!H_n^{(s)})}{2^{n+1}}$$
where $B_n(x_1,\dots,x_n)$ or denoted $Y_n(x_1,\dots,x_n)$ is the $...
- 3,577
4
votes
0
answers
184
views
On the generating function $\sum_{n=1}^\infty\left(e^{H_n}\log\left(H_n\right)\right)x^n$, for $0<x<1$
You can read Lagarias equivalence to the Riemann's Hypothesis from this MathWorld, I am saying (5), or well from the reference [1].
Let for real numbers $0<x<1$ the factor $x^n$, where $n\geq 1$...
user243301
2
votes
1
answer
122
views
Sum of Harmonic Number error over all $n$
It is well known that
$\displaystyle \mathcal{H}_n - \log{\left(n+\frac{1}{2}\right)} - \gamma = \frac{1}{24n^2} -\frac{1}{24n^3} + \mathcal{O}\left(\frac{1}{n^4}\right)$
Thus I pose the following ...
- 4,534
14
votes
2
answers
582
views
Generalized Harmonic Number Summation $ \sum_{n=1}^{\infty} {2^{-n}}{(H_{n}^{(2)})^2}$
Prove That $$ \sum_{n=1}^{\infty} \dfrac{(H_{n}^{(2)})^2}{2^n} = \tfrac{1}{360}\pi^4 - \tfrac16\pi^2\ln^22 + \tfrac16\ln^42 + 2\mathrm{Li}_4(\tfrac12) + \zeta(3)\ln2 $$
Notation : $ \displaystyle H_{n}...
- 5,616
3
votes
1
answer
259
views
Harmonic Generating Function
I have noticed an interesting generating function involving Harmonic Numbers.
$$\sum_{n=1}^{\infty}H_nx^n=\frac{\ln(1-x)}{x-1}$$
But, I have not seen a generating function involving second-order ...
- 463
13
votes
4
answers
744
views
Identity with Harmonic and Catalan numbers
Can anyone help me with this.
Prove that
$$2\log \left(\sum_{n=0}^{\infty}\binom{2n}{n}\frac{x^n}{n+1}\right)=\sum_{n=1}^{\infty}\binom{2n}{n}\left(H_{2n-1}-H_n\right)\frac{x^n}{n}$$
Where $H_n=\sum_{...
- 251
1
vote
2
answers
638
views
generating function for harmonic sequence
How I can find the generating function for the sequence
$$ \frac{ H_n } n $$
where $H_n$ -harmonic numbers.
I know that
$$
\sum\limits_{k=1}^{\infty}H_kz^k = -\frac{\ln(1 - z)}{1 - z}
$$
So, what ...
- 519
13
votes
3
answers
2k
views
Harmonic Numbers series I
Can it be shown that
\begin{align}
\sum_{n=1}^{\infty} \binom{2n}{n} \ \frac{H_{n+1}}{n+1} \ \left(\frac{3}{16}\right)^{n} = \frac{5}{3} + \frac{8}{3} \ \ln 2 - \frac{8}{3} \ \ln 3
\end{align}
where $...
- 26.8k