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Tagged with harmonic-numbers alternative-proof
32 questions
4
votes
1
answer
157
views
Simpler proof of identity $ \sum_{n=1}^{\infty} \frac{H_n^2}{n2^n} = \frac{7}{8} \zeta(3) $
The identity in question can be obtained by first proving
$$ \sum_{n=1}^{\infty} \frac{H_n^2}{n} z^n = - \frac{1}{3} \log^3(1-z) -\log(1-z) \text{Li}_2(z) + \text{Li}_3(z), \hspace{0.5cm} |z|<1, $$
...
- 1,773
17
votes
3
answers
671
views
Evaluating $\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x$
I was able to find
$$\int_0^{\frac{\pi}{2}}x^2 \cot x\ln(1-\sin x)\mathrm{d}x=-\frac14\sum_{n=1}^\infty\frac{4^n}{{2n\choose n}}\frac{H_{2n}}{n^3}$$
$$=5\operatorname{Li}_4\left(\frac12\right)-\frac{...
- 26.6k
4
votes
1
answer
341
views
Closed form for $\sum_{k=1}^\infty\frac{H_k^{(m)}}{k^n}$
Let's define
$$\sigma(m,n)=\sum_{k=1}^\infty\frac{H_k^{(m)}}{k^n}$$
where $H_k^{(m)}=\sum_{n=1}^{k}\frac{1}{n^m}$ is the k-th generalized harmonic number of order $m$.
In mathworld site Eq (20), I ...
- 26.6k
7
votes
0
answers
114
views
Convergence of double integral involving logarithms and binomials? Can anyone present a simpler proof?
I am working through a paper I posted to arXiv and am looking to condense my results as well as seek alternative proofs for some of my intermediate results. In particular, I am looking for a shorter ...
- 6,279
9
votes
1
answer
334
views
Different ways to evaluate $\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$
The following question:
How to compute the harmonic series $$\sum_{n=1}^\infty\frac{H_nH_n^{(2)}}{(n+1)(n+2)(n+3)}$$
where $H_n=\sum_{k=1}^n\frac{1}{k}$ and $H_n^{(2)}=\sum_{k=1}^n\frac{1}{k^2}$, was ...
- 26.6k
6
votes
1
answer
337
views
Trying to prove $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\frac1{2^2}+\frac1{3^2}+\frac1{4^2}+\cdots\right)$
It has been more than 7 days I have been trying to prove this following result using Harmonic Numbers
Let me add this
Proving $\left(1-\frac13+\frac15-\frac17+\cdots\right)^2=\frac38\left(\frac1{1^2}+\...
- 1,201
4
votes
1
answer
372
views
Prove $\displaystyle\sum_{-\infty}^0\text H(x)=\frac12\sum_0^\infty\frac{\text H_\frac x2-\text H_\frac{x-1}2}{x!}=\frac{\text{Ei}(2)-\text{Ei}(1)}e$
The Hadamard Gamma function generalizes the factorial function to have no poles as a result of the reciprocal gamma function in its definition. It also is monotonic on $[-\infty,0]$ and $[2,\infty]$ ...
- 12.8k
9
votes
4
answers
348
views
Alternative ways to evaluate $\int_0^1 x^{2n-1}\ln(1+x)dx =\frac{H_{2n}-H_n}{2n}$
For all, $n\geq 1$, prove that
$$\int_0^1 x^{2n-1}\ln(1+x)\,dx =\frac{H_{2n}-H_n}{2n}\tag1$$
This identity I came across to know from here, YouTube which is proved in elementary way.
Trying to make ...
- 3,482
4
votes
2
answers
349
views
How to approach $\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}$ elegantly?
How to show that
$$\sum_{n=0}^\infty(-1)^n\frac{H_{2n+1}}{(2n+1)^3}=\frac{\psi^{(3)}\left(\frac14\right)}{384}-\frac{\pi^4}{48}-\frac{35\pi}{128}\zeta(3)$$
without using the generating function:
\...
- 26.6k
3
votes
0
answers
95
views
Identity for $\gamma$ and proving the identity in different ways .
Following is the well know result that connects Harmonic number with binomial coefficients,
$$\color{red}{H_n}=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k}{n\choose k}$$
This motivates me and I came to ...
- 3,482
6
votes
2
answers
386
views
Computing $\sum_{n=1}^\infty\frac{(-1)^nH_n^2}{2n+1}$ in an alternative way
The following equality
$$\sum_{n=1}^\infty(-1)^{n-1}\frac{H_n^2}{2n+1}=\frac{3}{16}\pi^3-\frac34\ln^2(2)\pi-8\Im\left\{\text{Li}_3\left(\frac{1+i}{2}\right)\right\}$$
can be proved if we are allowed ...
- 26.6k
3
votes
3
answers
216
views
Alternative proof of computing $\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}$
In this solution we showed that
$$\sum _{n=1}^{\infty } \frac{4^n H_n}{n^2 {2n\choose n}}=6\ln(2)\zeta(2)+\frac72\zeta(3)\tag1$$
using the identity
$$\frac{\arcsin x}{\sqrt{1-x^2}}=\sum_{n=1}^\infty\...
- 26.6k
8
votes
3
answers
243
views
Evalute triple sum $ \sum_{m\geq 0}\sum_{n \geq 0}\sum_{p\geq 0} \frac{m!n!p!}{(m+n+p+2)!}$
I had a triple $T$ sum to evaluate
$$\sum_{m\geq 0}\sum_{n \geq 0}\sum_{p\geq 0} \frac{m!n!p!}{(m+n+p+2)!}$$ where $!$ denotes factorials.
The managed to find the closed form of it, $\displaystyle \...
- 3,482
4
votes
0
answers
82
views
How to show $\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}$ using series manipulation?
Using integration, I managed to show that
$$\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{H_{n-1}}{n^2(n+k)^2}=\sum_{n=1}^\infty\frac{H_{n-1}^{(2)}}{n^3}\tag1$$
But I would like to prove the equality using ...
- 26.6k
1
vote
0
answers
147
views
Different approach to compute $\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{1+x}2\right)\ dx$
The following integral $$I=\int_0^1\frac{\ln(x)}{1+x}\text{Li}_2\left(\frac{1+x}2\right)\ dx$$
was already evaluated by @Knas here where he found
$$I=-2\operatorname{Li}_4\left(\dfrac{1}{2}\right)-\...
- 26.6k
2
votes
4
answers
278
views
Compute $\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$
How to prove that
$$\int_0^1\frac{\ln(1-x)\ln(1+x)}{1+x}\ln\left(\frac{1+x}{2}\right)\ dx$$
$$=2\text{Li}_4\left(\frac12\right)-2\zeta(4)+\frac{15}8\ln(2)\zeta(3)-\frac12\ln^2(2)\zeta(2)$$
where $\...
- 26.6k
3
votes
0
answers
201
views
Evaluating $\int_0^1\frac{\ln x\operatorname{Li}_2(x)\ln(1-x/2)}{x}\ dx$
How to evaluate
$$\int_0^1\frac{\ln x\operatorname{Li}_2(x)\ln(1-\frac{x}{2})}{x}\ dx\ ?$$
I came across this integral while I was trying to calculate $\int_0^{1/2}\frac{\operatorname{Li}_2^2(x)}{...
- 26.6k
5
votes
0
answers
151
views
Computing $\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^4}$
I managed to find
$$S=\sum_{n=1}^\infty\frac{(-1)^{n-1}H_n^{(2)}}{(2n+1)^4}=20\beta(6)+\frac12\zeta(2)\beta(4)-\frac{7\pi^3}{32}\zeta(3)-\frac{31\pi}{8}\zeta(5)$$
where $\beta(a)$ is the ...
- 26.6k
2
votes
2
answers
182
views
Computing $\sum_{n=1}^\infty(-1)^n\frac{\overline{H}_nH_n}{n^2}$
How to elegantly prove that
$$S=\sum_{n=1}^\infty(-1)^n\frac{\overline{H}_nH_n}{n^2}=3\operatorname{Li}_4\left(\frac12\right)-\frac{29}{16}\zeta(4)-\frac34\ln^22\zeta(2)+\frac18\ln^42$$
where $\...
- 26.6k
10
votes
2
answers
453
views
Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=-\frac5{16}\zeta(3)$
How to prove that
$$S=\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^3=-\frac5{16}\zeta(3)$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
I came up ...
- 26.6k
4
votes
2
answers
358
views
Prove $\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}$
how to prove that
$$\sum_{n=0}^\infty(-1)^n(\overline{H}_n-\ln2)^2=\frac{\pi^2}{24}\ ?$$
where $\overline{H}_n=\sum_{k=1}^n\frac{(-1)^{k-1}}{k}$ is the alternating harmonic number.
This problem ...
- 26.6k
2
votes
2
answers
566
views
Compute $\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx$
How to prove
$$\int_0^1\frac{\ln^2(1+x)\operatorname{Li}_2(-x)}{x}dx=4\operatorname{Li}_5\left(\frac12\right)+4\ln2\operatorname{Li}_4\left(\frac12\right)-\frac{125}{32}\zeta(5)-\frac{1}{8}\zeta(2)...
- 26.6k
14
votes
2
answers
511
views
An attempt to prove the generalization of $\sum_{n=1}^\infty \frac{(-1)^nH_n}{n^{2a}}$
The following classical generalization
$$\sum_{n=1}^\infty\frac{(-1)^{n}H_n}{n^{2a}}=-\left(a+\frac 12\right)\eta(2a+1)+\frac12\zeta(2a+1)+\sum_{j=1}^{a-1}\eta(2j)\zeta(2a+1-2j)$$
where $\eta(a)=\...
- 26.6k
6
votes
2
answers
483
views
How to compute $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}$ by real integration only?
How to prove, by real methods that
$$\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^4}=\frac12\zeta(2)\zeta(3)-\frac{59}{32}\zeta(5)$$
where $H_n$ is the harmonic number and $\zeta$ is the Riemann zeta ...
- 26.6k
4
votes
0
answers
215
views
Compute $\int_0^1\frac{\tan^{-1}(x)\ln(1+x^2)}{x(1+x)}dx$ using harmonic series
How to prove
$$\mathcal{I}=\int_0^1\frac{\tan^{-1}(x)\ln(1+x^2)}{x(1+x)}dx=\frac{\pi^3}{96}-\frac{\pi}{8}\ln^2(2)$$
This problem is proposed by a friend and I managed to compute it using only ...
- 26.6k
1
vote
2
answers
219
views
How to compute $\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}$?
How to prove
$$\sum_{n=1}^\infty\frac{(-1)^nH_{n/2}}{n^4}=\frac18\zeta(2)\zeta(3)-\frac{25}{32}\zeta(5)?$$
I came across this series while I was working on a nice integral $\int_0^1\frac{\ln(1+x)\...
- 26.6k
12
votes
5
answers
720
views
Prove $\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$
How to prove
$$\int_0^1\frac{x^{2n}}{1+x}dx=\ln2+H_n-H_{2n}$$
I used this identity to solve some advanced harmonic series but I didn't provide a proof so I see that it's worth a post so that we ...
- 26.6k
3
votes
1
answer
378
views
Advanced : Compute $\sum_{n=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{n^5}$
How to prove the following equality
$$\mathcal S=\sum_{n=1}^\infty\frac{H_n^4-6H_n^2H_n^{(2)}+8H_nH_n^{(3)}+3\left(H_n^{(2)}\right)^2-6H_n^{(4)}}{n^5}\\=672\zeta(9)-240\zeta(2)\zeta(7)-105\zeta(3)\...
- 26.6k
8
votes
1
answer
697
views
Resistant integral $\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$
Prove, without using harmonic series, that
$$I=\int_0^1\left(\frac{\ln^2(1-x)\ln^2(1+x)}{1-x}-\frac{\ln^2(2)\ln^2(1-x)}{1-x}\right)\ dx$$
$$=\frac18\zeta(5)-\frac12\ln2\zeta(4)+2\ln^22\zeta(3)-\...
- 26.6k
6
votes
3
answers
741
views
How to find $\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}$ using real analysis and in an elegant way?
I have already evaluated this sum:
\begin{equation*}
\sum_{n=1}^{\infty}\frac{H_nH_{2n}}{n^2}=4\operatorname{Li_4}\left( \frac12\right)+\frac{13}{8}\zeta(4)+\frac72\ln2\zeta(3)-\ln^22\zeta(2)+\frac16\...
- 26.6k
1
vote
1
answer
256
views
Can we prove the divergence of harmonic series like this?
Let $S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\cdots$.
Or $S=(1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots)+(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}...
- 2,481
5
votes
6
answers
607
views
$\sum_{k=1}^nH_k = (n+1)H_n-n$. Why?
This is motivated by my answer to this question.
The Wikipedia entry on harmonic numbers gives the following identity:
$$
\sum_{k=1}^nH_k=(n+1)H_n-n
$$
Why is this?
Note that I don't just ...
- 25.4k