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$\sum_{n=1}^{\infty}=\frac{1}{16n^2-8n-3} \Rightarrow s_n=\sum_{k=1}^{n}\frac{1}{16k^2-8k-3}$

$\frac{1}{16k^2-8k-3}=\frac{1}{(4k-3)(4k+1)}=\frac{1}{4}\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)$

$\Rightarrow \sum_{k=1}^{n}\frac{1}{4}\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)=\frac{1}{4}\left(1-\frac{1}{4n+1}\right) \ \ \ (*)$

$\Rightarrow S=\lim_{n \to \infty} \frac{1}{4}\left(1-\frac{1}{4n+1}\right)=\frac{1}{4}$

The question is about the (*) line. Why does the first member equal the second one?

Thank you.

  • 1
    See http://en.wikipedia.org/wiki/Telescoping_series2012-01-09

2 Answers 2

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$ \eqalign { \sum_{k=1}^n( {1\over 4k-3} -{1\over 4k+1}) } $ is a "telescoping series". Write it out explicitly, and note most of the terms cancel. $ \eqalign {\textstyle \sum\limits_{k=1}^n( {1\over 4k-3} \!-\!{1\over 4k+1})&=\textstyle ( {1\over 1} \color{maroon}{-{1\over 5}})\! +\! (\color{maroon}{1\over5} \color{darkgreen}{\!-\!{1\over 9}})\!+\! (\color{darkgreen}{1\over 9} {\! -\!{1\over 13}}) \!+\!\cdots\! +\!({1\over 4n-7} \color{maroon}{\!-\!{1\over 4n-3}} ) +\!(\color{maroon}{1\over 4n-3}\! -\!{1\over 4n+1} )\cr &=\textstyle{1-{1\over 4n+1}}. } $

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$\sum_{k=1}^n(\frac{1}{4k-3}-\frac{1}{4k+1})$ $=(\frac{1}{1}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{9})+\dots+(\frac{1}{4n-7}-\frac{1}{4n-3})+(\frac{1}{4n-3}-\frac{1}{4n+1})$ $=\frac{1}{1}-(\frac{1}{5}-\frac{1}{5})-(\frac{1}{9}-\frac{1}{9})-\dots-(\frac{1}{4n-3}-\frac{1}{4n-3})-\frac{1}{4n+1}$ $=1-\frac{1}{4n+1}$

The (*) line should be $\frac{1}{4}(1-\frac{1}{4n+1})$, not $\frac{1}{4}(1-\frac{1}{4k+1})$