Question

Find the sum of the infinite series $1+\frac{1}{|\underset{–}{2}}+\frac{1}{|\underset{–}{4}}+\frac{1}{|\underset{–}{6}}+...$

Answer 1

$e^x=1+x+\frac{1}{|\underset{–}{2}}{x}^2+\frac{1}{|\underset{–}{3}}{x}^3+\frac{1}{|\underset{–}{4}}{x}^4+...$;

put $x=1$

We have $e=1+1+\frac{1}{|\underset{–}{2}}+\frac{1}{|\underset{–}{3}}+\frac{1}{|\underset{–}{4}}+...$;
and by putting $x=-1$ in the series for $e^x$
$e^{-1}=1-1+\frac{1}{|\underset{–}{2}}-\frac{1}{|\underset{–}{3}}+\frac{1}{|\underset{–}{4}}-...$
$\therefore e+e^{-1}=2(1+\frac{1}{|\underset{–}{2}}+\frac{1}{|\underset{–}{4}}+\frac{1}{|\underset{–}{6}}+...)$;
hence the sum of the series is $\frac{1}{2}(e+e^{-1})$.