Convergence of $\sum_{n=1}^{\infty}e^{-\sqrt{n}}$ using the integral test

Question

Given the series : $$\sum_{n=1}^{\infty}e^{-\sqrt{n}}$$ Determine if convergent or divergent.

The function is positive and monotonically decreasing function so I've used the "Integral Test"

$$\int_{1}^{\infty}e^{-\sqrt{x}}$$

then:$$ e^{-\sqrt{x}}\leq e^{\sqrt{x}}\leq e^{x}$$

$$\int_{1}^{\infty}e^{x}dx$$ which is clearly Divergent but the answer for some reason is convergent

Edit: Sorry for the misinterpretation. I forgot that if $f(x)>g(x)$ and $f(x)$ is divergent, it doesn't necessarily mean that $g(x)$ is divergent(contrary to convergent). Does anyone have a general direction for how do I suppose to solve it? I think that integral test is the most natural direction for solving this

Answer 1

Hint:

Near $+\infty$, $\;n^2=o\bigl(\mathrm e^{\sqrt n}\bigr)$, whence $\;\mathrm e^{-\sqrt n}=o\Bigl(\dfrac1{n^2}\Bigr)$.

Answer 2

For each $n\in\mathbb N$, $e^{-\sqrt n}\leqslant\sqrt ne^{-\sqrt n}$. But the series $\displaystyle\sum_{n=1}^\infty\sqrt ne^{-\sqrt n}$ converges, by the integral test:$$\int_1^\infty\sqrt xe^{-\sqrt x}=\lim_{M\to\infty}\bigl[-\left(2x+4\sqrt x+4\right)e^{-\sqrt x}\bigr]_1^M=\frac 6e.$$Therefore, your series converges, too.

Answer 3

$u=\sqrt x;\ du=\frac{1}{2\sqrt x}dx\Rightarrow \int_{1}^{\infty}e^{-\sqrt{x}}dx=2\int_{1}^{\infty}ue^{-u}du $ and this integral converges.

Answer 4

Hint.

$$ \frac{1}{e^{\sqrt n}}\le \frac{1}{e^{1.2\ln n}} = \frac{1}{n^{1.2}} $$

Answer 5

You're confusing the sandwich/squeeze theorem with the integral test.

The sandwich theorem says that if $g_1(x)<f(x)<g_2(x)$ and $g_1$ and $g_2$ converge, then $f$ converges. The converse does not hold: if $g_1$ and $g_2$ diverge, that does not mean that $f$ diverges. If the converse were true, then any function that is smaller than $e^x$ would be divergent, which is absurd.

The integral test, on the other hand, is an "if and only if" test, but that just takes the integral of the sequence. See https://en.wikipedia.org/wiki/Integral_test_for_convergence . So you should take just the integral of $e^{\sqrt x}$, not $e^{-\sqrt x}$ or $e^{x}$

Answer 6

I usually think that Integral Test is the least elegant way to show the convergence of a sequence. I do have to admit that it is convenient, simple and powerful, though.

In order to show convergence of series involving negative power of $e$, using the Taylor Expansion of $e^x$ is a good way.

Notice that

$$e^\sqrt{n} = \sum_{k=0}^\infty \frac{(\sqrt{n})^k}{k!} > \frac{n^2}{4!}$$

we have

$$ \sum_{n=1}^\infty e^{-\sqrt{n}} = \sum_{n=1}^\infty \frac{1}{e^\sqrt{n}} < \sum_{n=1}^\infty \frac{4!}{n^2} $$

and then you should be able to prove that the series converges.