The numerators diverge much faster this time

Calculus Level 3

It is well known that

e=k=11(k1)!=10!+11!+12!+13!+ e = \sum_{k=1}^\infty \dfrac{1}{(k-1)!} = \dfrac{1}{ 0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots

This value is also known (see Calvin's problem)

k=1k(k1)!=10!+21!+32!+43!+=x \sum_{k=1}^{\infty} \dfrac{ k} {(k-1)!} = \dfrac{1}{0!} + \dfrac{2}{ 1!} + \dfrac{3}{2!} + \dfrac{ 4}{3!} + \cdots \quad = x (xx is kept secret here so Calvin's problem isn't spoiled.)

So what is

k=1k2(k1)!=10!+41!+92!+163!+? \sum_{k=1}^{\infty} \dfrac{k^2} {(k-1)!} = \dfrac{1}{0!} + \dfrac{4}{ 1!} + \dfrac{9}{2!} + \dfrac{16}{3!} + \cdots \quad ?

Notation: !! denotes the factorial notation. For example, 8!=1×2×3××88! = 1\times2\times3\times\cdots\times8 . And 0!=10!=1 as always.

×

Problem Loading...

Note Loading...

Set Loading...