비교 판정법

As usual, let's see whether the “easy" tests work. The terms we're adding converge to zero: $$\lim_{n \to \infty}\frac{e^{-\sqrt{n}}}{\sqrt{n}}=\lim_{n \to \infty}\frac{1}{\sqrt{n}e^{\sqrt{n}}}=0$$ so the divergence test is inconclusive. Our series isn't geometric, and it doesn't seem obvious how to compare it to a geometric series. However, the terms we're adding seem like they would make an integrable function.

Set $f(x)=\frac{e^{-\sqrt{x}}}{\sqrt{x}}$. For $x\ge1$, this function is positive and decreasing (since it is the product of the two positive decreasing functions $e^{-\sqrt x}$ and $\frac1{\sqrt x}$). We use the integral test with this function. Using the substitution $u=\sqrt x$, so that $\mathrm{d}u = \frac1{2\sqrt x}\,\mathrm{d}x$, we see that

$$\begin{aligned} \int_1^\infty f(x)\,\mathrm{d}x &=\lim_{R\rightarrow\infty}\int_1^R \bigg( \frac{e^{-\sqrt{x}}}{\sqrt{x}}\,\mathrm{d}x \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( \int_1^{\sqrt{R}} e^{-u}\cdot2\,\mathrm{d}u \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( {-}2e^{-u}\Big|_1^{\sqrt{R}} \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( {-}2e^{-\sqrt R} + 2e^{-\sqrt1} \bigg) = 0 + 2e^{-1}, \end{aligned}$$

and so this improper integral converges. By the integral test, the given series also converges.