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For any positive a and n, it seems this inequality holds

$$ \sum\limits_{t=n+1}^\infty e^{-at} \leq \frac{1}{a}e^{-an} $$

How can I prove this inequality and does this holds for negative a ?

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  • $\begingroup$ For $a > 0$ you can compute the left-hand side explicitly (using the geometric series formula). $\endgroup$ – Martin R Aug 16 at 11:06
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If $a \le 0$, the left-hand sum doesn't converge.

If $a > 0$, then your inequality is true. Using the geometric series formula, we have $$\sum_{t=n+1}^{\infty}e^{-at} = \frac{e^{-a(n+1)}}{1- e^{-a}} = e^{-an} \frac{e^{-a}}{1-e^{-a}}.$$

This less than or equal to $\frac{1}{a}e^{-an}$, since $ \frac{e^{-a}}{1-e^{-a}}\le \frac{1}{a}$. To see this, recall that $a \le \frac{1 - e^{-a}}{e^{-a}} = e^a - 1$.

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