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How to calculate this sum $$\sum_{n=1}^{\infty} \frac{H_n \cdot H_{n+1}}{(n+1)(n+2)}$$

Attempt

The series telescopes. We have $$=\frac{H_n \cdot H_{n+1}}{(n+1)(n+2)} = \frac{H_n \cdot H_{n+1}}{n+1} - \frac{H_n \cdot H_{n+1}}{n+2}$$

$$=\frac{H_n(H_n + \frac{1}{n+1})}{n+1} - \frac{(H_{n+1} - \frac{1}{n+1}) \cdot H_{n+1}}{n+2}$$

$$=\frac{H_n^2}{n + 1} - \frac{H_{n+1}^2}{n + 2} + \frac{H_n}{(n + 1)^2}+ \frac{H_{n+1}}{(n + 1)(n + 2)}$$

$$=\frac{H_n^2}{n + 1} - \frac{H_{n+1}^2}{n + 2} + \frac{H_{n+1} - \frac{1}{n+1}}{(n+1)^2} + - \frac{H_{n+1}}{n+1} - \frac{H_{n+1}}{n+2}$$

$$=\frac{H_n^2}{n + 1} - \frac{H_{n+1}^2}{n + 2} + \frac{H_{n+1}}{(n+1)^2} - \frac{1}{(n+1)^3} + \frac{H_{n + 1}}{n + 1} - \frac{H_{n + 2}}{n + 2} + \frac{1}{{(n + 2)^2}}$$

It follows that $$\sum_{n=1}^{\infty} \frac{H_n \cdot H_{n+1}}{(n+1)(n+2)} = \sum_{n=1}^{\infty} \left(\frac{H_{n}^2}{n + 1} - \frac{H_{n+1}^2}{(n+2)^2}\right) + \sum_{n=1}^{\infty} \frac{H_{n+1}}{(n+1)^2} - \sum_{n=1}^{\infty} \frac{1}{(n+1)^3}$$

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  • $\begingroup$ @Derivative Isn't it $a_n \sim \frac{\log^2 n}{n^2}$? $\endgroup$
    – user
    Commented Feb 11 at 18:05
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    $\begingroup$ For what it's worth, Mathematica gives for the sum: $\frac{\pi^2}{6}+\zeta(3)$. $\endgroup$
    – Hans Olo
    Commented Feb 11 at 18:15
  • $\begingroup$ @I disagree with Hans Olo. I get $2.607..$ rather than $\zeta(2) + \zeta(3) = 2.84699..$ $\endgroup$
    – mick
    Commented Feb 11 at 19:20
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    $\begingroup$ @mick Yes, I want to know how to progress further and I’m open to different methods. $\endgroup$
    – user1285841
    Commented Feb 11 at 19:28
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    $\begingroup$ @HansOlo I am sorry, you are correct , I used alpha and he did not spend enough time to get the correct value... +1 $\endgroup$
    – mick
    Commented Feb 11 at 19:37

2 Answers 2

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We use the following summation by parts formula (twice): $$\sum_{n=1}^\infty a_{n+1}(b_{n+1}-b_n) = -a_1 b_1 - \sum_{n=1}^\infty (a_{n+1}-a_n)b_n \tag1\label1$$ First take $a_n=H_{n-1}H_n$ and $b_n=-1/(n+1)$ in \eqref{1} to obtain \begin{align} \sum_{n=1}^\infty \frac{H_n H_{n+1}}{(n+1)(n+2)} &= \sum_{n=1}^\infty H_n H_{n+1}\left(\frac{1}{n+1}-\frac{1}{n+2}\right) \\ &= \frac{-H_0 H_1}{-(n+1)} - \sum_{n=1}^\infty (H_n H_{n+1}-H_{n-1}H_n)\frac{-1}{n+1} \\ &= 0 + \sum_{n=1}^\infty \frac{H_n (H_{n+1}-H_{n-1})}{n+1} \\ &= \sum_{n=1}^\infty \frac{H_n}{n+1}\left(\frac{1}{n+1}+\frac{1}{n}\right) \\ &= \sum_{n=1}^\infty \frac{H_n}{(n+1)^2} + \sum_{n=1}^\infty \frac{H_n}{n(n+1)} \\ &= \zeta(3) + \sum_{n=1}^\infty H_n\left(\frac{1}{n}-\frac{1}{n+1}\right). \end{align} Now take $a_n=H_{n-1}$ and $b_n=-1/n$ in \eqref{1} to obtain $$\zeta(3) + \frac{-H_0}{-1} + \sum_{n=1}^\infty (H_n-H_{n-1})\frac{1}{n} =\zeta(3) + 0 + \sum_{n=1}^\infty \frac{1}{n^2} = \zeta(3) + \frac{\pi^2}{6}.$$

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  • $\begingroup$ Weird. Mathematica gave a different answer. Let me try again ... $\endgroup$
    – mick
    Commented Feb 11 at 19:31
  • $\begingroup$ wolframalpha.com/… This gives the smaller 2.607 ? $\endgroup$
    – mick
    Commented Feb 11 at 19:33
  • $\begingroup$ maybe due to a timeout error ? $\endgroup$
    – mick
    Commented Feb 11 at 19:34
  • $\begingroup$ I tried Mathematica and got the same: $\zeta(3) + \frac{\pi^2}{6}$ $\endgroup$
    – jjagmath
    Commented Feb 11 at 19:34
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    $\begingroup$ @mick Click it again. :) $\endgroup$
    – RobPratt
    Commented Feb 11 at 19:37
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Multiply both sides of $$\sum_{n=1}^\infty\frac{H_{n}}{n+1}x^{n+1}=\frac12\ln^2(1-x)\tag{1}$$ by $-\ln(1-x)$ then integrate from $0$ to $1$ using $-\int_0^1 x^{n+1}\ln(1-x)dx=\frac{H_{n+2}}{n+2}$, we get

$$\sum_{n=1}^{\infty} \frac{H_{n} \cdot H_{n+2}}{(n+1)(n+2)}=-\frac12\int_0^1\ln^3(1-x)dx=3\tag{2}$$

Now multiply $(1)$ by $-\ln(x)$ then integrate using $-\int_0^1 x^{n+1}\ln(x)dx=\frac{1}{(n+2)^2}$, we get

$$\sum_{n=1}^{\infty} \frac{H_{n}}{(n+1)(n+2)^2}=-\frac12\int_0^1 \ln(x)\ln^2(1-x)dx=3-\zeta(2)-\zeta(3)\tag{3}$$

Take the difference of $(2)$ and $(3)$, we find

$$\zeta(2)+\zeta(3)=\sum_{n=1}^{\infty} \frac{H_{n}}{(n+1)(n+2)}\left(H_{n+2}-\frac{1}{n+2}\right)=\sum_{n=1}^{\infty} \frac{H_{n} \cdot H_{n+1}}{(n+1)(n+2)}$$

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    $\begingroup$ Really impressive!!… (+1) $\endgroup$
    – user1285841
    Commented Feb 14 at 17:39
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    $\begingroup$ (+1) Nice job, Ali! :-) $\endgroup$
    – user97357329
    Commented Feb 14 at 18:46

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