# Sum of exponential function inequality

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 ?

• For $a > 0$ you can compute the left-hand side explicitly (using the geometric series formula). – Martin R Aug 16 at 11:06

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 $$a \le \frac{1 - e^{-a}}{e^{-a}} = e^a - 1$$.