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First, the original problem follows, $$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}$$ where $$H_{k}=\sum_{j=1}^k \frac{1}{j}$$ is the $k$-th partial sum of harmonic series.

Using the following identity, $$H_{k+1} = \sum_{j=1}^\infty (\frac{1}{j} - \frac{1}{k+j+1}).$$ I was able to get this one. $$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}=\sum_{j=1}^\infty [\frac{1}{j} - \frac{1}{j+1} + \frac{1}{j+1}\sum_{k=1}^\infty \frac{1}{(k+1)(k+1+j)}] $$ So, If I get the sum of the double series, $$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}.$$ I can also find the original problem.

What method can I use at this problem?

Robert Z
  • 145,942
Yoongiable
  • 63

4 Answers4

5

Going back to the original problem, note that for any positive integer $N$, $$ \begin{align} \sum_{k=1}^N \frac{H_{k+1}}{k(k+1)}&=\sum_{k=1}^N \frac{H_{k+1}}{k}-\sum_{k=1}^N \frac{H_{k+1}}{k+1}\\ &=\sum_{k=1}^N \frac{H_{k}}{k}+\sum_{k=1}^N \frac{1}{k(k+1)}-\sum_{k=2}^{N+1} \frac{H_{k}}{k}\\ &=1+\sum_{k=1}^N \left(\frac{1}{k}-\frac{1}{k+1}\right)- \frac{H_{N+1}}{N+1}\\ &=2-\frac{1}{N+1}- \frac{H_{N+1}}{N+1}. \end{align}$$ Hence, by taking the limit as $N\to +\infty$, we get $$\sum_{k=1}^{\infty} \frac{H_{k+1}}{k(k+1)}=2.$$

P.S. It follows that $$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}=1.$$

Robert Z
  • 145,942
3

We show here how to solve the original problem.

Partial fraction decomposition gives

$$\frac{1}{k (k+1)}=\frac{1}{k}-\frac{1}{k+1}\tag{1}$$

Hence the sum can be written as

$$s = \sum _{k=1}^{\infty } \left(\frac{1}{k}-\frac{1}{k+1}\right) H_{k+1}$$

$$=\sum _{k=1}^{\infty } \left(\frac{H_{k+1}}{k}-\frac{H_{k+1}}{k+1}\right)$$

Now we have

$$H_{k+1}=H_{k}+\frac{1}{k+1}$$

so that $s=s_{1} + s_{2}$ where

$$ \begin{align} s_{1}&=\sum _{k=1}^{\infty } \left(\frac{H_k}{k}-\frac{H_{k+1}}{k+1}\right)\\ &=( \frac{H_1}{1}-\frac{H_2}{2})+(\frac{H_2}{2}-\frac{H_3}{3}) +\text{...} =\frac{H_1}{1}=1\\ s_{2}&=\sum _{k=1}^{\infty } \frac{1}{k (k+1)}=\sum _{k=1}^{\infty } \left(\frac{1}{k}-\frac{1}{k+1}\right)\\ & =( \frac{1}{1}-\frac{1}{2})+(\frac{1}{2}-\frac{1}{3}) +\text{...} =\frac{1}{1}=1 \end{align}$$

As we can see both series telescope so that we get $s_{1}=1$ and $s_{2}=1$

so that finally $s=2$.

Dr. Wolfgang Hintze
  • 12,465
1

As an alternative approach, $$S=\sum_{j,k\geq 1}\frac{1}{(j+1)(k+1)(j+k+1)}=\sum_{j,k\geq 1}\iiint_{(0,1)^3}x^j y^k z^{j+k}\,dx\,dy\,dz$$ and Fubini's theorem allows to switch the double series and the triple integral (this sounds a bit ludicrous, I know), leading to $$\begin{eqnarray*} S &=&\iiint_{(0,1)^3}\frac{xyz^2}{(1-xz)(1-yz)}\,dx\,dy\,dz\\&=&\int_{0}^{1}\iint_{(0,z)^2}\frac{XY}{(1-X)(1-Y)}\,dX\,dY\,\frac{dz}{z^2}\\ &=&\int_{0}^{1}\left[\int_{0}^{z}\frac{w}{1-w}\,dw\right]^2\,\frac{dz}{z^2} \\&=&\int_{0}^{1}\left[1+\frac{\log(1-z)}{z}\right]^2\,dz\\&=&1-2\,\zeta(2)+\int_{0}^{1}\frac{\log^2(1-z)}{z^2}\,dz\\&=&1-2\,\zeta(2)+\int_{0}^{1}\frac{\log^2(z)}{(1-z)^2}\,dz\\&=&1-2\,\zeta(2)+\sum_{m\geq 1}\int_{0}^{1}m z^{m-1}\log^2(z)\,dz\\&=&1-2\,\zeta(2)+2\sum_{m\geq 1}\frac{1}{m^2}=\color{red}{\large 1}.\end{eqnarray*}$$

Jack D'Aurizio
  • 353,855
0

$$\frac{H_{k+1}}{k(k+1)}=\frac{H_{k+1}}k-\frac{H_{k+1}}{k+1}=\frac{H_k}k+\frac1k-\frac1{k+1}-\frac{H_{k+1}}{k+1}$$ nicely telescopes to the limit

$$\frac{H_1}1+\frac11.$$