A harmonic sum with 'k'

Question

I am stuck at this , Any help would be appreciated

$ \displaystyle \sum_{n=1}^{\infty} {\rm H}_{n}(\frac{1}{n}-\frac{1}{n+k}) $

How do we calculate this ?

Answer 1

According to Lemma 1 in this article by A. Sofo and D. Cvijovic, we have that $$\sum_{n=1}^{\infty} H_{n}\left(\frac{1}{n}-\frac{1}{n+k}\right)= k\sum_{n=1}^{\infty} \frac{H_{n}}{n(n+k)}= \frac{\pi^2}{6}+\frac{H^2_{k-1}+H^{(2)}_{k-1}}{2}$$ where $H_n=\sum_{j=1}^n\frac{1}{j}$ and $H^{(2)}_n=\sum_{j=1}^n\frac{1}{j^2}$.

Answer 2

Some ideas here : Write it as $$ \sum_{n=0}^\infty \sum_{i=1}^n \frac{1}{i\ n} \ \ - \sum_{n=0}^\infty \sum_{i=1}^n \frac{1}{i(n+k)} $$ Now let's explicitly write some terms of the positive sum and of the negative sum in a triangle-shape: $$ \frac{1}{1}\frac{1}{1} + \\ \frac{1}{1}\frac{1}{2} + \frac{1}{2} \frac{1}{2} + \\ \frac{1}{1} \frac{1}{3} + \frac{1}{2} \frac{1}{3} + \frac{1}{3} \frac{1}{3} + \\ ... $$ $$ - ( \\ \frac{1}{1}\frac{1}{1+k} + \\ \frac{1}{1}\frac{1}{2+k} + \frac{1}{2} \frac{1}{2+k} + \\ \frac{1}{1} \frac{1}{3+k} + \frac{1}{2} \frac{1}{3+k} + \frac{1}{3+k} \frac{1}{3} + \\ ... ) $$

Let's call the $j$th diagonal of such a triangle the "line" that starts from the $j$'th row at the most left term and goes down "diagonally" to the bottom right corner. For example the $1$-diagonal starts at the top of the triangle and goes through all the terms on the right side.

We see that the sum of these two triangles will cancel leaving only the $k$'th first diagonals (that is all the term in diagonal $1,2,...,k$) of the "positive triangle" : $$ \sum_{i=1}^k \sum_{j=1}^\infty \frac{1}{j(j+i-1)} $$ Where the outer sum selects the diagonal and the inner sum goes through all the elements. In particular we can extract the first sum where $i=1$ like so : $$ \sum_{j=1}^\infty \frac{1}{j^2} + \sum_{i=2}^k \sum_{j=1}^\infty \frac{1}{j(j+i-1)} = \frac{\pi^2}{6} + \sum_{i=2}^k \sum_{j=1}^\infty \frac{1}{j(j+i-1)} $$ My mathematical knowledge stops me here because Mathematica yields a too complicated expression for the rest of the sum. Hope it helps a bit :)

Answer 3

We have: $$ -\log(1-x) = \sum_{n\geq 1}\frac{x^n}{n}\tag{1} $$ hence: $$ \frac{-\log(1-x)}{x(1-x)} = \sum_{n\geq 1} H_n x^{n-1} \tag{2} $$ and: $$ \sum_{n\geq 1}H_n\left(\frac{1}{n}-\frac{1}{n+k}\right) = \int_{0}^{1}\frac{-\log(1-x)}{x(1-x)}(1-x^k)\,dx \tag{3}$$ can be easily computed by expanding $\frac{1-x^k}{1-x}$ as a geometric sum, then exploiting: $$ \int_{0}^{1}-\log(1-x)x^{m-1}\,dx = \frac{H_m}{m}\tag{4} $$ and: $$ \sum_{m=1}^{k}\frac{H_m}{m}=\sum_{m=1}^{k}\frac{1}{m}\sum_{1\leq s\leq m}\frac{1}{s} = \frac{H_k^2+H_k^{(2)}}{2}.\tag{5} $$ Putting everything together, $$\boxed{\,\forall k\in\mathbb{N}^+,\qquad \sum_{n\geq 1}\frac{k H_n}{n(n+k)}=\color{red}{\zeta(2)+\frac{H_{k-1}^2+H_{k-1}^{(2)}}{2}}\,}\tag{6} $$