## Calculus (3rd Edition)

The series $\sum_{n=3}^{\infty}\frac{ 1}{e^{\sqrt n} }$ converges.
We have the series: $\sum_{n=3}^{\infty}\frac{ 1}{e^{\sqrt n} }$ We apply the integral test: $\int_1^{\infty} e^{-\sqrt{x}}dx=\int_1^{\infty} 2ue^{-u}du$ Where $u=\sqrt{x}$ We solve by integration by parts: $\int_1^{\infty} 2ue^{-u}du=-2e^{-u}(1+u)|_1^{\infty}=\lim_{R\rightarrow\infty} -2e^{-u}(1+u)|_1^{R}=\lim_{R\rightarrow\infty}-2e^{-R}(1+R)-[-2e^{-1}(1+1)]=\lim_{R\rightarrow\infty}\frac{-2(1+R)}{e^R}+\frac{4}{e}$ This leads to an indeterminate form, so we apply L'Hopital's rule: $=\lim_{R\rightarrow\infty}\frac{-2}{e^R}=0$ Since the integral converges, then the series converges by the integral test.