# Why does $\sum\limits_{k=1}^{n}\frac{1}{4}(\frac{1}{4k-3}-\frac{1}{4k+1})$ turn into $\frac{1}{4}(1-\frac{1}{4n+1})$?

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

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

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

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

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

Thank you.

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The sum is of the form $(a_1 - a_2) + (a_2 - a_3) + \dots + (a_{n-1} - a_{n})$; terms cancel in pairs except for the first and last, leaving $a_1 - a_{n}$. – mjqxxxx Jan 9 '12 at 15:19
– Thomas Andrews Jan 9 '12 at 15:31

\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}}. }
$$\sum_{k=1}^n(\frac{1}{4k-3}-\frac{1}{4k+1})$$ $$=(\frac{1}{1}-\frac{1}{5})+(\frac{1}{5}-\frac{1}{9})+\dots+(\frac{1}{4n-7}-\frac{1}{4n-3})+(\frac{1}{4n-3}-\frac{1}{4n+1})$$ $$=\frac{1}{1}-(\frac{1}{5}-\frac{1}{5})-(\frac{1}{9}-\frac{1}{9})-\dots-(\frac{1}{4n-3}-\frac{1}{4n-3})-\frac{1}{4n+1}$$ $$=1-\frac{1}{4n+1}$$
The (*) line should be $\frac{1}{4}(1-\frac{1}{4n+1})$, not $\frac{1}{4}(1-\frac{1}{4k+1})$