How find this limits $\lim_{N\to\infty}\sum_{n=1}^{N}\frac{1}{(n+1)}\sum_{i=1}^{n}\frac{1}{i(n+1-i)}$

Question

Find this limits

$$\lim_{N\to\infty}\sum_{n=1}^{N}\left(\dfrac{1}{(n+1)}\left(\dfrac{1}{1\cdot n}+\dfrac{1}{2\cdot(n-1)}+\cdots+\dfrac{1}{(n-1)\cdot 2}+\dfrac{1}{n\cdot 1}\right)\right)$$

I know this $$\dfrac{1}{1\cdot n}+\dfrac{1}{2\cdot(n-1)}+\cdots+\dfrac{1}{(n-1)\cdot 2}+\dfrac{1}{n\cdot 1}=\sum_{i=i}^{n}\dfrac{1}{i(n+1-i)}=\dfrac{1}{n+1}\sum_{i=1}^{n}\left(\dfrac{1}{i}+\dfrac{1}{n+1-i}\right)=\dfrac{2H_{n}}{n+1}$$ so this limits is $$\sum_{n=1}^{\infty}\dfrac{2H_{n}}{(n+1)^2}$$ then I can't,Thank you

Answer 1

I will prove that

$$\sum_{n=1}^\infty \frac{H_n}{n^2} = 2\zeta(3)$$

Your sum can be obtained by manipulating the indices.

First use that

$$H_n = \int^1_0 \frac{1-x^n}{1-x}\,dx$$

$$\int^1_0 \frac{1}{1-x}\sum_{n = 1}^\infty\frac{1-x^n }{n^2}\,dx = \int^1_0 \frac{\zeta(2)-\mathrm{Li}_2(x)}{1-x}dx$$

Now use the duplication formula for dilogarithm

$$\zeta(2)-\mathrm{Li}_2(x) = \mathrm{Li}_2(1-x)+\log(x)\log(1-x)$$

Hence we have

$$\int^1_0 \frac{\mathrm{Li}_2(1-x)+\log(x)\log(1-x)}{1-x}dx = \int^1_0 \frac{\mathrm{Li}_2(x)+\log(1-x)\log(x)}{x}dx$$

$$\int^1_0 \frac{\mathrm{Li}_2(x)}{x} = \mathrm{Li}_3(1) =\zeta(3)$$

$$\int^1_0 \frac{\log(1-x)\log(x)}{x}dx =\int^1_0 \frac{\mathrm{Li}_2(x)}{x} = \zeta(3) $$

$$\sum_{n=1}^\infty \frac{H_n}{n^2} = 2 \zeta(3)$$

Another approach

we have the following general formula

$$\tag{1} \sum_{k\geq 1} \frac{H_k}{k^2} \, x^{k} = \text{Li}_3(x)-\, \text{Li}_3(1-x)+\, \log(1-x) \text{Li}_2(1-x)\\ +\frac{1}{2}\log(x) \log^2(1-x)+\zeta(3)$$

Let $x = 1$ to get the desired formula.

(1) can be derived using the approach in this thread.

Answer 2

$\displaystyle \begin{align} \sum_{n=1}^{\infty}\dfrac{H_n}{(n+1)^2} &= \sum_{n=1}^{\infty}\left(\dfrac{H_{n+1}}{(n+1)^2} - \frac{1}{(n+1)^3}\right) \\ &= \sum_{n=2}^{\infty}\left(\dfrac{H_{n}}{n^2} - \frac{1}{n^3}\right) \\ &= \sum_{n=1}^{\infty}\left(\dfrac{H_{n}}{n^2} - \frac{1}{n^3}\right) \\ &= 2\zeta(3) - \zeta(3)\end{align}$

The last sum $\displaystyle \sum_{n=1}^{\infty}\dfrac{H_{n}}{n^2} = 2\zeta(3)$ is proved here for the general case $\displaystyle \sum_{n=1}^{\infty}\dfrac{H_{n}}{n^q}$.