The harmonic numbers are given by $$H_n=\sum_{k=1}^n\frac{1}{k}.$$ Numerical calculation suggests $$ \sum_{k=1}^{n}(-1)^k{n\choose k}{n+k\choose k}\sum_{i=1}^{k}\frac{1}{n+i}=(-1)^nH_n. $$ I can not give a proof and want to know if this identity exists in the literature. If not, how can we prove it?

Hints, references or proof are all welcome.

The harmonic numbers are given by $$H_n=\sum_{k=1}^n\frac{1}{k}.$$ Numerical calculation suggests $$ \sum_{k=1}^{n}(-1)^k{n\choose k}{n+k\choose k}\sum_{i=1}^{k}\frac{1}{n+i}=(-1)^nH_n. $$ I can not give a proof of this identity. How to prove it?

Hints, references or proof are all welcome.

The harmonic numbers are given by $$H_n=\sum_{k=1}^n\frac{1}{k}.$$ Then we have $$ \sum_{k=1}^{n}(-1)^k{n\choose k}{n+k\choose k}\sum_{i=1}^{k}\frac{1}{n+i}=(-1)^nH_n. $$ I want to know if this identity exists in the literature. If not, how can we prove it?

Hints, references or proof are all welcome.

The harmonic numbers are given by $$H_n=\sum_{k=1}^n\frac{1}{k}.$$ Numerical calculation suggests $$ \sum_{k=1}^{n}(-1)^k{n\choose k}{n+k\choose k}\sum_{i=1}^{k}\frac{1}{n+i}=(-1)^nH_n. $$ I can not give a proof and want to know if this identity exists in the literature. If not, how can we prove it?

Hints, references or proof are all welcome.