Sum of Harmonic Numbers

Question

Similar to this question , let $H_n$ be the $n^{th}$ harmonic number,

$$ H_n = \sum_{i=1}^{n} \frac{1}{i}$$

Is there a similar method to calculate the following?:

$$\sum_{i=1}^{n}iH_i$$

Answer 1

I am ready to bet that Jack D'Aurizio will provide one of his elegant answers.

Just mimicing what he answered in the previous question, by analogy with $$\int x\log(x) dx=\frac{1}{2} x^2 \log (x)-\frac{x^2}{4}$$ you have $$\sum_{i=1}^niH_i=\frac{1}{4} n (n+1) \left(2 H_{n+1}-1\right)$$

Answer 2

The most straightforward method to calculating this sum is probably just to change the order of summation:

$$\begin{align} \sum_{i=1}^{n}iH_{i} &=\sum_{i=1}^{n}\sum_{k=1}^{i}\frac{i}{k}\\ &=\sum_{k=1}^{n}\sum_{i=k}^{n}\frac{i}{k}\\ &=\sum_{k=1}^{n}\frac{(1+n-k)(k+n)}{2k}\\ &=\sum_{k=1}^{n}\frac{n+n^2+k-k^2}{2k}\\ &=\frac{n+n^2}{2}\sum_{k=1}^{n}\frac{1}{k}+\frac12\sum_{k=1}^{n}(1-k)\\ &=\frac{n(1+n)}{2}H_n+\frac{n(1-n)}{4}. \end{align}$$