How to write it formally? Convergence of an infinite sum

Question

I know that the following sum $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{1+2n}} e^{-\sqrt{n}}$$ converges because it's basically looking at $$\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}} e^{-\sqrt{n}} \sim \int_{1}^{\infty} \frac{1}{\sqrt{x}} e^{-\sqrt{x}} dx = \frac{2}{e},$$ which is convergent. But how can I make this reasoning more formal?

Answer 1

With your results as given you could use the limit comparison test. Observe that $$ \frac{\left(\displaystyle \frac{1}{\sqrt{1+2n}} e^{-\sqrt{n}}\right)}{\left(\displaystyle \frac{1}{\sqrt{n}} e^{-\sqrt{n}}\right)} = \sqrt{ \frac{n}{1+2n} } \xrightarrow{n \to \infty} \frac{1}{\sqrt 2} \in (0,\infty) $$ Since you know that the simpler sum converges, you have the convergence of the desired sum.

Answer 2

$$ \sum_{n=1}^{\infty}\frac{e^{-\sqrt{n}}}{\sqrt{1+2n}}\le \sum_{n=1}^{\infty}\frac{e^{-\sqrt{n}}}{\sqrt{n}}\le e^{-1}+\int_{1}^{\infty} \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ dx=\frac{3}{e} $$

Answer 3

$$\begin{align*}\sum_{n=1}^{\infty} \frac{e^{-\sqrt{n}}}{\sqrt{1+2n}}&<\sum_{n=1}^{\infty} \frac{e^{-\sqrt{n}}}{\sqrt{2n}} \\ &\sim\sum_{n=1}^{\infty} \frac{e^{-\sqrt{n}}}{\sqrt{n}}\end{align*}$$ $e^{-\sqrt n}/\sqrt{n}$ is continuous and monotone decreasing on $n\in[1,\infty)$, hence by the integral test, $$\sum_{n=1}^\infty\frac{e^{-\sqrt n}}{\sqrt{n}}\sim\underbrace{\int_1^\infty\frac{e^{-\sqrt x}}{\sqrt x}\ dx}_{2/e}$$ the improper integral is finite and the series converges.