Extrakt04

2022-06-28

Why does $\sum _{k=1}^{n}\frac{1}{4}\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)$ turn into $\frac{1}{4}\left(1-\frac{1}{4n+1}\right)$

hofyonlines5

Expert

$\begin{array}{r}\sum _{k=1}^{n}\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)\end{array}$
is a "telescoping series". Write it out explicitly, and note most of the terms cancel.
$\begin{array}{rl}\sum _{k=1}^{n}\left(\frac{1}{4k-3}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}\frac{1}{4k+1}\right)& =\left(\frac{1}{1}-\frac{1}{5}\right)\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{5}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}\frac{1}{9}\right)\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{9}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}\frac{1}{13}\right)\phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\cdots \phantom{\rule{negativethinmathspace}{0ex}}+\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{4n-7}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}\frac{1}{4n-3}\right)+\phantom{\rule{negativethinmathspace}{0ex}}\left(\frac{1}{4n-3}\phantom{\rule{negativethinmathspace}{0ex}}-\phantom{\rule{negativethinmathspace}{0ex}}\frac{1}{4n+1}\right)\\ & =1-\frac{1}{4n+1}.\end{array}$

Cory Patrick

Expert

$\sum _{k=1}^{n}\left(\frac{1}{4k-3}-\frac{1}{4k+1}\right)$
$=\left(\frac{1}{1}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{9}\right)+\cdots +\left(\frac{1}{4n-7}-\frac{1}{4n-3}\right)+\left(\frac{1}{4n-3}-\frac{1}{4n+1}\right)$
$=\frac{1}{1}-\left(\frac{1}{5}-\frac{1}{5}\right)-\left(\frac{1}{9}-\frac{1}{9}\right)-\cdots -\left(\frac{1}{4n-3}-\frac{1}{4n-3}\right)-\frac{1}{4n+1}$
$=1-\frac{1}{4n+1}$
The (*) line should be $\frac{1}{4}\left(1-\frac{1}{4n+1}\right)$, not $\frac{1}{4}\left(1-\frac{1}{4k+1}\right)$

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