An Iterated Sum Using Calculus And/Or Algebra

$\displaystyle f(n) = \sum_{k=2}^{\infty} \frac{1}{k^nk!}$

Thus,

$\displaystyle \sum_{n=2}^{\infty} f(n) = \sum_{n=2}^{\infty} \sum_{k=2}^{\infty} \frac{1}{k^nk!}$

Switching the sums,

$\displaystyle \sum_{n=2}^{\infty} f(n) = \sum_{k=2}^{\infty} \sum_{n=2}^{\infty} \frac{1}{k^nk!}$

$\displaystyle \sum_{n=2}^{\infty} f(n) = \sum_{k=2}^{\infty} \frac{1}{k!} \sum_{n=2}^{\infty} \frac{1}{k^n}$

Now,
$\displaystyle \sum_{n=2}^{\infty} \frac{1}{k^n} = \frac{1}{k^2} +\frac{1}{k^3} + \ldots = \frac{\frac{1}{k^2}}{1-\frac{1}{k}} = \frac{1}{k(k-1)}$

Hence,

$\displaystyle \sum_{n=2}^{\infty} f(n) = \sum_{k=2}^{\infty} \frac{1}{k!}\times \frac{1}{k(k-1)}$

$\displaystyle \sum_{n=2}^{\infty} f(n) = \sum_{k=2}^{\infty} \frac{1}{k^2(k-1)(k-1)!}$

$\displaystyle \sum_{n=2}^{\infty} f(n) = \sum_{k=2}^{\infty} \frac{1}{(k-1)!} \left(\frac{1}{k-1}-\frac{k+1}{k^2}\right)$

$\displaystyle \sum_{n=2}^{\infty} f(n) = \sum_{k=2}^{\infty} \frac{1}{(k-1)!} \left(\frac{1}{k-1} -\frac{1}{k} -\frac{1}{k^2}\right)$

$\displaystyle \sum_{n=2}^{\infty} f(n) = \sum_{k=2}^{\infty} \left(\frac{1}{(k-1)(k-1)!}-\frac{1}{kk!}-\frac{1}{k!}\right)$

Note that the first two terms form a Telescoping series , thus,

$\displaystyle \sum_{n=2}^{\infty} f(n) = 1 - \sum_{k=2}^{\infty}\frac{1}{k!}$

Now,
$\displaystyle \sum_{k=2}^{\infty} \frac{1}{k!} = e - 1 - 1 = e-2$

Hence,

$\displaystyle \sum_{n=2}^{\infty} f(n) = 1- (e-2) = 3 - e = \boxed{0.2817}$

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

changing the position of sum

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

$(\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ k! } { \left( \frac { 1 }{ { k }^{ 2 } } +\frac { 1 }{ { k }^{ 3 } } +.....+\infty \right) } } )$

$(\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ k! } { \left( 1+\frac { 1 }{ k } +\frac { 1 }{ { k }^{ 2 } } +\frac { 1 }{ { k }^{ 3 } } +.....+\infty \quad -1-\frac { 1 }{ k } \right) } } )$

$\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ \left( k-1 \right) *\left( k-1 \right) ! } -\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ k! } } } -\sum _{ k=2 }^{ \infty }{ \frac { 1 }{ k*k } } =1-e+2$