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

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

• You replaced your integral with something much larger. You can't conclude anything about the convergence or divergence of your integral. – Michael Burr Oct 8 '18 at 22:08
• You say, about your original function "The function is positive and monotonically decreasing", and then compare it to a monotonically increasing function $e^x$? Why? – Yuriy S Oct 9 '18 at 8:15
• @YuriyS thanks for the feedback! – ProgC Oct 9 '18 at 11:37
• @ProgC, you are welcome, hope the answers in this thread were helpful – Yuriy S Oct 9 '18 at 13:47

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

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:$$\int_1^\infty\sqrt xe^{-\sqrt x}=\lim_{M\to\infty}\left[-\left(2x+4\sqrt x+4\right)e^{-\sqrt x}\right]_1^M=\frac 6e.$$Therefore, your series converges, too.

Hint.

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

Hint:

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

• what those little zeros mean? – James Oct 8 '18 at 22:18
• It's little oh (not $0$), used in asymptotic analysis. Roughly speaking, $f=o(g)$ means the ratio $\dfrac fg$ tends to $0$ at $\infty$. – Bernard Oct 8 '18 at 22:21
• Ohhh but this To me it is too fancy and probably to the question guy too lol, see jose answer o matematleta answer – James Oct 8 '18 at 22:24
• Maybe it's fancy, but it's standard and very short. The result (not the notations, of course) is known from high school. – Bernard Oct 8 '18 at 22:28
• Perhaps it's worth knowing that the $2$ can be any $p>1$. As in, the series converges faster than any $p$-series does. – Kyle Miller Oct 9 '18 at 4:09

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

The sandwich theorem says that if $$g_1(x) and $$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 $$e^{x}$$

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.