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Find the sum of the following infinite series $$e^{-x}\sum_{i=0}^{\infty}\dfrac{i.x^i}{i!}$$

The summation looks like an exponential series but how to tackle that?$$ 0+\frac{x}{1!}+\frac{2x^2}{2!}+...$$

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$$\sum_{n=0}^{\infty}\dfrac{nx^n}{n!} = x\cdot\sum_{n=1}^{\infty}\dfrac{x^{n-1}}{(n-1)!} = xe^x.$$

So $e^{-x}$ times that is just $x$. Is this what you're asking?

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Note that $$ \sum_{i=0}^\infty \frac{ix^i}{i!} =xD(e^x)=xe^x. $$ So $$ e^{-x}\left(\sum_{i=0}^\infty \frac{ix^i}{i!}\right) =e^{-x}xe^x =x. $$

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