## 13th Harvard MIT Calculus problem 8

Let $f(n)=\sum\limits_{k=2}^{\infty }{\frac{1}{{{k}^{n}}k!}}$. Calculate $\sum\limits_{k=2}^{\infty }{f(n)}$ 😀

Solution

—> $\sum\limits_{k=2}^{\infty }{f(n)}=\sum\limits_{k=2}^{\infty }{\sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}k!}}}$

$=\sum\limits_{k=2}^{\infty }{\frac{1}{k!}\sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}}}}$

we know that $\sum\limits_{n=0}^{\infty }{\frac{1}{{{k}^{n}}}}=\frac{1}{{{k}^{0}}}+\frac{1}{{{k}^{1}}}+...........=\frac{1}{1-\frac{1}{k}}=\frac{k}{k-1}$

So $\sum\limits_{n=2}^{\infty }{\frac{1}{{{k}^{n}}}}=\frac{k}{k-1}-1-\frac{1}{k}=\frac{1}{k(k-1)}$

—> $\sum\limits_{n=2}^{\infty }{f(n)}=\sum\limits_{k=2}^{\infty }{\frac{1}{k!}.\frac{1}{k(k-1)}}$

$=\sum\limits_{k=2}^{\infty }{\frac{1}{(k-1)!}}.\frac{1}{{{k}^{2}}(k-1)}$

–>  $=\sum\limits_{k=2}^{\infty }{\frac{1}{(k-1)!}\left( \frac{1}{k-1}-\frac{1}{{{k}^{2}}}-\frac{1}{k} \right)}$

$=\sum\limits_{k=2}^{\infty }{\left( \frac{1}{(k-1)!(k-1)}-\frac{1}{k.k!}-\frac{1}{k!} \right)}$

$=\sum\limits_{k=2}^{\infty }{\left( \frac{1}{(k-1)!(k-1)}-\frac{1}{k.k!} \right)}-\sum\limits_{k=2}^{\infty }{\frac{1}{k!}}$

*** By Taylor series $\sum\limits_{k=0}^{\infty }{\frac{1}{k!}}=e$

Therefore $\sum\limits_{n=2}^{\infty }{f(n)=\frac{1}{1\cdot 1!}-\left( e-\frac{1}{0!}-\frac{1}{1!} \right)=3-e}$