Summation with fractions, discrete calculus

Question

Here is the summation question

$$\sum_{k=1}^{n-1}k\left(1+\frac{1}{2}+\dots+\frac{1}{k}\right)$$

I think it should be solved by the technique of discrete calculus (summation by parts). Can someone give me a hint or show me how to do this?

Answer 1

By noting that $k=\binom{k+1}{2}-\binom{k}{2}$ and by using summation by parts, we find that $$\sum_{k=1}^{n-1}kH_k=\sum_{k=1}^{n-1}\left(\binom{k+1}{2}H_k-\binom{k}{2}\left(H_{k-1}+\frac{1}{k}\right)\right)=\binom{n}{2}\left(H_n-\frac{1}{2}\right)$$ where $H_k=1+\frac{1}{2}+\dots+\frac{1}{k}$. Can you fill the details and show the above identity?

Answer 2

The elementary way:

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

$$\sum_{k=1}^n H_k=1+(1+\frac{1}{2})+(1+\frac{1}{2}+\frac{1}{3})+....+(1+\frac{1}{2}+\frac{1}{2}+...+\frac{1}{n})= \sum_{k=1}^{n} \frac{n-(k-1)}{k}$$ $$=(n+1)H_n-n$$ Next $$(k+1)^2H_{k+1}-k^2H_k=2kH_k+H_k+k+1$$ By telescopic summing $$\frac{1}{2}[(n+1)^2 H_{n+1}-1]=\sum_{k=1}^{n} k H_k +\frac{1}{2}\sum_{k=1}^n H_k +\frac{1}{2}\sum_{k=1}^n (k+1)$$ $$=\sum_{k=1}^{n} k H_k +\frac{1}{2}[(n+1) H_n-n]+\frac{1}{4}(n^2+3n)$$ So the required sum is $$\implies \sum_{k=1}^{n} k H_k= \frac{1}{2}(n+1)^2 H_{n+1}-\frac{(n+1)H_n}{2}-\frac{n^2+n+2}{4}$$ OR $$\sum_{k=1}^{n-1} k H_k=\frac{1}{2}n^2H_n-\frac{1}{2}n H_{n-1}-\frac{n^2-n+2}{4}=\frac{1}{2} n(n-1) ~ [H_n-\frac{1}{2}].$$