\rm\sum_(k = 1)^ \infty (1)/(k(k + n)) = (H_n)/(n) \\​

Question

`\rm\sum_(k = 1)^ \infty (1)/(k(k + n)) = (H_n)/(n) \\`​

Answer 1

Pretty straightforward. Partial fractions, then observe the tail ends of the two subsequent sums cancel each other.

`\displaystyle \sum_(k=1)^\infty \frac1{k(k+n)} = \frac1n \sum_(k=1)^\infty \left(\frac1k - \frac1{k+n}\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \frac1n \left(\sum_(k=1)^\infty \frac1k - \sum_(k=1)^\infty \frac1{k+n}\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \frac1n \left(\sum_(k=1)^\infty \frac1k - \sum_(k=n+1)^\infty \frac1k\right) \\\\ ~~~~~~~~~~~~~~~~~~ = \frac1n \sum_(k=1)^n \frac1k \\\\ ~~~~~~~~~~~~~~~~~~ = \frac{H_n}n`

where

`\displaystyle \sum_(k=1)^n \frac1k = H_n`

is the
`n`-th harmonic number.