Linked Questions

138 votes
16 answers
10k views

Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$

Let $$A(p,q) = \sum_{k=1}^{\infty} \frac{(-1)^{k+1}H^{(p)}_k}{k^q},$$ where $H^{(p)}_n = \sum_{i=1}^n i^{-p}$, the $n$th $p$-harmonic number. The $A(p,q)$'s are known as alternating Euler sums. ...
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131 votes
10 answers
29k views

How to find ${\large\int}_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx$

Please help me to find a closed form for this integral: $$I=\int_0^1\frac{\ln^3(1+x)\ln x}x\mathrm dx\tag1$$ I suspect it might exist because there are similar integrals having closed forms: $$\begin{...
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80 votes
8 answers
8k 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 ...
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48 votes
6 answers
2k views

A series involving the harmonic numbers : $\sum_{n=1}^{\infty}\frac{H_n}{n^3}$

Let $H_{n}$ be the nth harmonic number defined by $ H_{n} := \sum_{k=1}^{n} \frac{1}{k}$. How would you prove that $$\sum_{n=1}^{\infty}\frac{H_n}{n^3}=\frac{\pi^4}{72}?$$ Simply replacing $H_{n}$ ...
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  • 2,203
56 votes
6 answers
2k views

Infinite Series $\sum\limits_{n=1}^\infty\frac{(H_n)^2}{n^3}$

How to prove that $$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$ $H_n$ denotes the harmonic numbers.
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32 votes
5 answers
2k views

Infinite Series $\sum_{n=1}^\infty\frac{H_n}{n^22^n}$

How can I prove that $$\sum_{n=1}^{\infty}\frac{H_n}{n^2 2^n}=\zeta(3)-\frac{1}{2}\log(2)\zeta(2).$$ Can anyone help me please?
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36 votes
6 answers
2k views

Find the closed form of $\sum_{n=1}^{\infty} \frac{H_{ n}}{2^nn^4}$

One of the possible ways of computing the series is to obtain the generating function, but this might be a tedious, hard work, pretty hard to obtain. What would you propose then? $$\sum_{n=1}^{\...
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  • 43.2k
57 votes
3 answers
2k views

Conjectured value of a harmonic sum $\sum_{n=1}^\infty\left(H_n-\,2H_{2n}+H_{4n}\right)^2$

There is a known asymptotic expansion of harmonic numbers $H_n$ for $n\to\infty$: $$\begin{align}H_n&=\gamma+\ln n+\sum_{k=1}^\infty\left(-\frac{B_k}{k\cdot n^k}\right)\\ &=\gamma+\ln n+\frac1{...
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26 votes
7 answers
2k views

Alternating harmonic sum $\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k$

How to analytically prove $$\sum_{k\geq 1}\frac{(-1)^k}{k^3}H_k=-\frac{11\pi^4}{360}+\frac{\ln^42-\pi^2\ln^22}{12}+2\mathrm{Li}_4\left(\frac12\right)+\frac{7\ln 2}{4}\zeta(3) $$ As O.L answer ...
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29 votes
4 answers
8k views

Calculating alternating Euler sums of odd powers

Definition $$\mathbf{H}_{m}^{(n)}(x) = \sum_{k=1}^\infty \frac{H_k^{(n)}}{k^m} x^k\tag{1}$$ We define $$\mathbf{H}_{m}^{(1)}(x) = \mathbf{H}_{m}(x)=\sum_{k=1}^\infty \frac{H_k}{k^m} x^k \tag{2}$$ ...
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33 votes
3 answers
3k views

Integral ${\large\int}_0^1\ln(1-x)\ln(1+x)\ln^2x\,dx$

This problem was posted at I&S a week ago, and no attempts to solve it have been posted there yet. It looks very alluring, so I decided to repost it here: Prove: $$\int_0^1\ln(1-x)\ln(1+x)\ln^...
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26 votes
3 answers
899 views

How prove this sum $\sum_{n=1}^{\infty}\binom{2n}{n}\frac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}$

show that $$\sum_{n=1}^{\infty}\binom{2n}{n}\dfrac{(-1)^{n-1}H_{n+1}}{4^n(n+1)}=5+4\sqrt{2}\left(\log{\dfrac{2\sqrt{2}}{1+\sqrt{2}}}-1\right)$$ where $H_{n}=1+\dfrac{1}{2}+\dfrac{1}{3}+\cdots+\dfrac{...
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26 votes
3 answers
976 views

Closed form for $\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n}$

Please help me to find a closed form for the sum $$\sum_{n=1}^\infty\frac{(-1)^n n^4 H_n}{2^n},$$ where $H_n$ are harmonic numbers: $$H_n=\sum_{k=1}^n\frac{1}{k}=\frac{\Gamma'(n+1)}{n!}+\gamma.$$
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14 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 $...
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  • 22.8k
21 votes
3 answers
1k views

Sum of Harmonic numbers $\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$

Finding the closed form of: $$\sum\limits_{n=1}^{\infty} \frac{H_n^{(2)}}{2^nn^2}$$ where, $\displaystyle H_n^{(2)} = \sum\limits_{k=1}^{n}\frac{1}{k^2}$ It appears when we try to determine the ...
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