I have the assumption that the following holds:
$$\lim_{n \to \infty}\frac{1}{n^2} \cdot \sum_{i = 0}^n \left(1 - \frac 1n \right)^i = 1 - \frac{2}{e}.$$
However, I am totally not sure about it. How can I prove or try to prove it?
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1Are you talking about the limit when $n \to \infty$ ? Otherwise, this is clearly false because $e$ is irrational. – lhf Oct 20 '14 at 15:57
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3Use the formula for a geometric sum. – Daniel Fischer Oct 20 '14 at 15:57
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1If you aren't talking about the limit, this is false, as it'd imply e is rational – Alan Oct 20 '14 at 15:58
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Yes, sorry, I forgot to add the limit. – Sacha Oct 21 '14 at 08:23
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As written, it is false. We have $$ \frac 1{n^2} \sum_{i=0}^n \left( 1 - \frac 1n \right)^i = \frac 1{n^2} \frac{1-(1-1/n)^{n+1}}{1-(1-1/n)} = \frac 1{n^2} \frac{1-(1-1/n)(1-1/n)^n}{1/n} = \frac{1-(1-1/n)(1-1/n)^n}n. $$ The top goes to $1-\frac 1e$, but the bottom goes to $\infty$. Perhaps you need to fix that $n^2$ to $n$ in your original expression to get $1-\frac 1e$, but you're not getting that $2$ anytime soon.
Hope that helps,
Patrick Da Silva
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