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- Feb 14, 2012

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- Thread starter anemone
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- Thread starter
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- #1

- Feb 14, 2012

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- Jan 26, 2012

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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 $S_n = \dfrac{1}{2} T_n$.