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I have an equation similar to the one below:

$$ \sum_{m=0}^{\infty} { mx^m\over m!} $$

which naturally reduces to:

$$ \sum_{m=0}^{\infty} {x^m\over (m-1)!} $$

and now for some reason we can do the following:

$$ = x \sum_{m=1}^{\infty} {x^{m-1}\over (m-1)!} $$

Why does the index increase? Is it just because we cannot evaluate $(m-1)!$ if $m=0$?

Note: if this works then we can shift the summation index to get that $ \sum_{m=0}^{\infty} { mx^m\over m!} = xe^x$.

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The first term is zero. $$ \sum_{m=0}^{\infty} \frac{mx^m}{m!} = \frac{0 x^0}{0!} + \sum_{m=1}^{\infty} \frac{m x^m}{m!} = \sum_{m=1}^{\infty} \frac{x^m}{(m-1)!} = x\sum_{m=1}^{\infty} \frac{x^{m-1}}{(m-1)!} = x\sum_{m=0}^{\infty} \frac{x^m}{m!}. $$

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Continuing your line of thought: $$\sum_{m=0}^{\infty} { mx^m\over m!}=\sum_{m=0}^{\infty} {x^m\over (m-1)!}=x\sum_{m=0}^{\infty} {x^{m-1}\over (m-1)!}=x\left(\frac{x^{-1}}{(-1)!}+\sum_{m=1}^{\infty} {x^{m-1}\over (m-1)!}\right)=\\ x\left(0+\sum_{m=0}^{\infty} {x^m\over m!}\right)=x\sum_{m=0}^{\infty} {x^m\over m!}.$$ Note: $\frac{1}{\Gamma{(0)}}=\frac1{(-1)!}=0$. Reference: Wikipedia.

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