Is $\sum_{t=1}^{\infty} e^{-\sqrt{t}}$ finite? Is $\displaystyle \sum_{t=1}^{\infty} e^{-\sqrt{t}}$ finite? This is not homework. Actually, a computer algebra package has told me it is finite, but I would like to see a proof. No approach worth trying has occurred to me - a hint would be appreciated...

Note that 
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots.$$ 
Put $x=\sqrt{t}$. We get that if $t\gt 0$ then
$$e^{\sqrt{t}}\gt \frac{t^2}{4!}.$$
Thus the $t$-th term of our series is $\lt \dfrac{24}{t^2}$.
The series $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}$ is convergent, and therefore so is $\displaystyle \sum_{n=1}^\infty \frac{24}{n^2}$. Thus by Comparison our series converges.
Remark: A very similar argument shows that if $\delta$ is any positive constant, then $\displaystyle\sum_{n=1}^\infty \frac{1}{e^{n^\delta}}$ converges.
