# The numerators diverge much faster this time

**Calculus**Level 4

It is well known that

\[ e = \sum_{k=1}^\infty \dfrac{1}{(k-1)!} = \dfrac{1}{ 0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots \]

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

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

So what is

\[ \sum_{k=1}^{\infty} \dfrac{k^2} {(k-1)!} = \dfrac{1}{0!} + \dfrac{4}{ 1!} + \dfrac{9}{2!} + \dfrac{16}{3!} + \cdots \quad ? \]

\[\] **Notation**: \(!\) denotes the factorial notation. For example, \(8! = 1\times2\times3\times\cdots\times8 \). And \(0!=1\) as always.

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