Compare S_n and T_n: Sums of Fractions

[anemone]

Compare \(\displaystyle S_n=\sum_{k=1}^{n}\frac{k}{(2n-2k+1)(2n-k+1)}\) and \(\displaystyle T_n=\sum_{k=1}^{n}\frac{1}{k}\).

 

[Mathwebster]

My solution

Spoiler

First we can re-write the sum as

$\displaystyle\sum_{k=1}^n \dfrac{1}{2n-2k+1} - \dfrac{1}{2n-k+1}$

Reversing the order of the sum gives

$\displaystyle\sum_{k=1}^n \dfrac{1}{2k-1} - \dfrac{1}{n+k}$

The first sum can be written as $T_{2n} - \dfrac{1}{2} T_n$ while the second $T_{2n} - T_n$. Simplify gives that $S_n = \dfrac{1}{2} T_n$.