May 19, 2013 Thread starter Admin #1 anemone MHB POTW Director Staff member Feb 14, 2012 3,835 Compare \(\displaystyle S_n=\sum_{k=1}^{n}\frac{k}{(2n-2k+1)(2n-k+1)}\) and \(\displaystyle T_n=\sum_{k=1}^{n}\frac{1}{k}\).
Compare \(\displaystyle S_n=\sum_{k=1}^{n}\frac{k}{(2n-2k+1)(2n-k+1)}\) and \(\displaystyle T_n=\sum_{k=1}^{n}\frac{1}{k}\).
May 19, 2013 #2 Jester Well-known member MHB Math Helper Jan 26, 2012 183 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 $S_n = \dfrac{1}{2} T_n$.
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 $S_n = \dfrac{1}{2} T_n$.