The Numerators Diverge Too

Calculus Level 2

It is well known that

$e = \sum_{k=0}^\infty \dfrac{1}{k!} = \dfrac{1}{ 0!} + \dfrac{1}{1!} + \dfrac{1}{2!} + \dfrac{1}{3!} + \cdots$

What is the value of

$\sum_{k=0}^{\infty} \dfrac{ k+1} {k!} = \dfrac{1}{0!} + \dfrac{2}{ 1!} + \dfrac{3}{2!} + \dfrac{ 4}{3!} + \cdots \quad ?$

Notation: $!$ denotes the factorial notation. For example, $8! = 1\times2\times3\times\cdots\times8$ .

Bonus: Prove this by just manipulating the terms from the first identity.

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