Series exercise asked by Hamza Mahmood in math group

Exercise:

Show that, $\sum\limits_{k=1}^{\infty }{\frac{k}{\left( 5+k \right)!}}$ is convergent

Solution: by using the ratio test this series is convergent

Let ${{a}_{k}}=\frac{k}{\left( 5+k \right)!}$ such that ${{a}_{k}}\ne 0$

So $\left| \frac{{{a}_{k+1}}}{{{a}_{k}}} \right|=\left| \frac{\frac{k+1}{\left( 5+k+1 \right)!}}{\frac{k}{\left( 5+k \right)!}} \right|=\left| \frac{\left( k+1 \right)\left( 5+k \right)!}{k\left( 6+k \right)!} \right|=\left| \frac{\left( k+1 \right)\left( 5+k \right)!}{k\left( 6+k \right)\left( 5+k \right)!} \right|=\left| \frac{k+1}{k\left( 6+k \right)} \right|$

So $L=\underset{k\to \infty }{\mathop{\lim }}\,\left| \frac{k+1}{k\left( 6+k \right)} \right|=\underset{k\to \infty }{\mathop{\lim }}\,\frac{1}{k}=0$  since $L=0<1$ so  $\sum\limits_{k=1}^{\infty }{\frac{k}{\left( 5+k \right)!}}$ is convergent

Let $u=5+k\Leftrightarrow k=u-5$

So $\sum\limits_{k=1}^{\infty }{\frac{k}{\left( 5+k \right)!}=\sum\limits_{u=6}^{\infty }{\frac{u-5}{u!}}=\sum\limits_{u=6}^{\infty }{\frac{u}{u!}}-\sum\limits_{u=6}^{\infty }{\frac{5}{u!}}=-5\sum\limits_{u=6}^{\infty }{\frac{1}{u!}+\sum\limits_{u=6}^{\infty }{\frac{u}{u\left( u-1 \right)!}}}}=-5\sum\limits_{u=6}^{\infty }{\frac{1}{u!}+\sum\limits_{u=6}^{\infty }{\frac{1}{\left( u-1 \right)!}}}$

Remember that ${{e}^{x}}=\sum\limits_{k=0}^{\infty }{\frac{{{x}^{k}}}{k!}}=1+x+\frac{{{x}^{2}}}{2!}+\frac{{{x}^{3}}}{3!}+\frac{{{x}^{4}}}{4!}+\frac{{{x}^{5}}}{5!}+\frac{{{x}^{6}}}{6!}+\frac{{{x}^{7}}}{7!}+....$

Put $x=1$ so $e=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{120}+\frac{1}{6!}+\frac{1}{7!}+....\Leftrightarrow e-\frac{163}{60}=\frac{1}{6!}+\frac{1}{7!}+\frac{1}{8!}+...$

Hence $5\sum\limits_{u=6}^{\infty }{\frac{1}{u!}=5\left( e-\frac{163}{60} \right)=5e-\frac{163}{12}}$

In similar way we get last term

$e=1+1+\frac{1}{2}+\frac{1}{6}+\frac{1}{24}+\frac{1}{5!}+\frac{1}{6!}+\frac{1}{7!}\Leftrightarrow e-\frac{65}{24}=\sum\limits_{u=6}^{\infty }{\frac{1}{\left( u-1 \right)!}}$

Thus $\sum\limits_{k=1}^{\infty }{\frac{k}{\left( 5+k \right)!}=-5e+\frac{163}{12}+e-\frac{65}{24}=-4e+\frac{87}{8}}$