# 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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