I'm stumped on how to show that $\sum_{k=1}^{\infty} k^2/k! = 2e$. We know that the RHS is $2 \sum_{k=0}^{\infty} \frac{1}{x^k}$, but I'm not sure how to show this the same as the LHS.
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The RHS is nothing like $2 \sum_{k=0}^{\infty} \frac{1}{x^k}$. Unless $x=1-1/e$. – TonyK Oct 02 '23 at 15:29