Question

Find the sum of the following infinite series:
$\sum _{n=0}^{\infty }\frac{1}{n!}\left[\sum _{k=0}^n(k+1){\int }_0^1{2}^{-(k+1)x}dx\right]$

Answer 1

${\int }_0^1{2^{-(k+1)x}dx=\left[\frac{2^{-(k+1)x}}{-(k+1)\log a}\right]}_0^1=\frac{2^{-(k+1)}-1}{-(k+1)\log a}$
$\therefore \sum _{k=0}^n(k+1).\left\{\frac{2^{-(k+1)x}}{-(k+1)\log a}\right\}$
$=\frac{1}{\log a}\sum _{k=0}^n\{1-\frac{1}{2^{k+1}}\}$
$=\frac{1}{\log a}\left[\right(n+1)-\{\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...(n+1\left)terms\right\}]$
$=\frac{1}{\log a}\left[\right(n+1)-\frac{\frac{1}{2}\{1-{\left(\frac{1}{2}\right)}^{n+1}\}}{(1-\frac{1}{2})}]$
$=\frac{1}{\log a}\left[\right(n+1)-(1-2^{-(n+1)}\left)\right]$
$=\frac{1}{\log a}[n+{\left(\frac{1}{2}\right)}^{n+1}]$
$\therefore S=\frac{1}{\log a}\sum _{n=0}^{\infty }[\frac{n}{n!}+\frac{1}{2}.\frac{{\left(\frac{1}{2}\right)}^n}{n!}]$
$=\frac{1}{\log a}\sum _{n=0}^{\infty }[\frac{1}{(n-1)!}+\frac{1}{2}.\frac{x^n}{n!}],$
where $x=\frac{1}{2}$
$=\frac{1}{\log a}[e+\frac{1}{2}{e}^x]=\frac{1}{\log a}(e+\frac{1}{2}\sqrt{e})$