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I was asked by a friend this problem: Show whether the series $$\sum_{n = 1}^{\infty} \int_{0}^{1 / \sqrt{n}} \frac{\mathrm{e}^{x} - 1}{1 + x} \mathop{}\!\mathrm{d} x$$ converges.

Now the integral does not have a closed form, and the summand evidently vanishes as $n \to \infty$. I've tried to apply all convergence tests that I could find, but none seem to work in this case. Many of them (d'Alembert, Dirichlet, etc.) try to evaluate $\displaystyle \frac{a_{n + 1}}{a_{n}}$ and then do something about it, but in this case, when the summand is a definite integral, this fraction doesn't seem to simply things much... Nor does it seem to simply things to evaluate the square root $\sqrt{\lvert a_{n}\rvert}$ or the integral $\displaystyle \int_{1}^{\infty} a_{n} \mathop{}\!\mathrm{d} n$. The Cauchy condensation test evaluates $2^{n} a_{2^{n}}$, which also doesn't seem to lead anywhere...

I think the reason the aforementioned tests don't seem to work (based on my trial anyway...maybe they do and I'm just not seeing it) is that the summand is an integral. Hence I was wondering if there is a convergence test which works for series with definite integral summand?

Edit: As Jack pointed out below, there is no need for a test specifically for series with integral summand. (It's techniques and tricks combined with available tests)

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    $\begingroup$ In this case, you can use the fact that $e^x \geq 1 + x$ for $x \geq 0$. $\endgroup$ – Michael Biro Jun 2 '17 at 2:09
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For $n >0$ , we have $$0\le x \le1$$

thus $$e^x-1\ge x $$ and $$\frac {1}{1+x}\geq \frac {1}{2}. $$

from this

$$\int_0^\frac {1}{\sqrt {n}}\frac {e^x-1}{1+x}dx\geq \frac {1}{2}\Bigl [\frac {x^2}{2}\Bigr]_0^\frac {1}{\sqrt {n}} $$

$$\geq \frac {1}{4n} $$

the series is Divergent.

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  • $\begingroup$ Thank you! I think your method is very clear. One point though: I think for $n \geq 1$ it's always that $0 \leq x \leq 1$ so the "$n$ enough great" part may not be necessary. Other than this tiny point your answer is perfect $\endgroup$ – akkuszaetifs Jun 2 '17 at 2:37
  • $\begingroup$ @akkuszaetifs Thanks for pointing that. $\endgroup$ – hamam_Abdallah Jun 2 '17 at 2:41
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A slightly less elementary, but more precise method.

Let $a_n\stackrel{\rm def}{=}\int_{0}^{1 / \sqrt{n}} \frac{\mathrm{e}^{x} - 1}{1 + x} \mathop{}\!\mathrm{d} x$, for $n\geq 1$.

By setting $u=\sqrt{n}x$, we have $$ na_n = n\int_{0}^{1 / \sqrt{n}} \frac{\mathrm{e}^{x} - 1}{1 + x} \mathop{}\!\mathrm{d} x = \sqrt{n}\int_0^1 \frac{e^{\frac{u}{\sqrt{n}}}-1}{1+\frac{u}{\sqrt{n}}}\!\mathrm{d} u $$ Defining $(f_n)_n$ on $[0,1]$ by $f_n(u) \stackrel{\rm def}{=}\sqrt{n}\frac{e^{\frac{u}{\sqrt{n}}}-1}{1+\frac{u}{\sqrt{n}}}$, it is easy to check that

  • $f_n(u_0) \xrightarrow[n\to\infty]{} u_0$ for every fixed $u_0\in[0,1]$ (pointwise convergence)
  • $\lvert f_n(u)\rvert \leq (e-1)u$ for all $u\in[0,1]$ and $n\geq 1$ (as $e^x - 1\leq (e-1)x$ for $x\in[0,1]$) (domination)

Thus, by the Dominated Convergence Theorem, we get $$ n a_n \xrightarrow[n\to\infty]{} \int_0^1 \lim_{n\to\infty} f_n(u)\, du = \int_0^1 u\, du = \frac{1}{2}$$ i.e. $$\boxed{a_n \operatorname*{\sim}_{n\to\infty} \frac{1}{2n}}$$

The divergence (and exact rate of divergence) of the series follow: $$\boxed{\sum_{n=1}^N a_n \operatorname*{\sim}_{n\to\infty} \sum_{n=1}^N \frac{1}{2n} \operatorname*{\sim}_{n\to\infty} \frac{1}{2}\ln N}$$

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Elementary inequalities do the job nicely: over the interval $[0,1]$ we have

$$\frac{e^x-1}{x+1}\geq \frac{x}{x+1} \geq x-x^2\tag{1}$$ and the series $\sum_{n\geq 1}\int_{0}^{1/\sqrt{n}}(x-x^2)\,dx$ is divergent by the p-test, so the original series is divergent as well.

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  • $\begingroup$ I put mine in the comments because I'm not sure it answers the actual question - namely, is there a convergence test which works for series with definite integral summand? $\endgroup$ – Michael Biro Jun 2 '17 at 2:14
  • $\begingroup$ @MichaelBiro:a definite integral is still a number, so why are you looking for different criteria? They are just the same. And they all derive from elementary inequalities, so when in doubt you may always "go back to the stone age". $\endgroup$ – Jack D'Aurizio Jun 2 '17 at 2:16
  • $\begingroup$ Yes, but OP asked the question of whether there is a test specifically designed for summing integrals, not for the answer to that specific sum. $\endgroup$ – Michael Biro Jun 2 '17 at 2:20
  • $\begingroup$ @MichaelBiro: in such a case I guess the answer is just no, because there is no need for such a thing. Definite integrals are just numbers. $\endgroup$ – Jack D'Aurizio Jun 2 '17 at 2:21
  • $\begingroup$ Thank you and @MichaelBiro for the critical insight! However, it doesn't really follow from $x \in {[0, 1]}$ that $1 - 1 / (x + 1) \geq 1 / 2$...? It seems to me that $0 \leq 1 - 1 / (x + 1) \leq 1 / 2$ for $x \in {[0, 1]}$. $\endgroup$ – akkuszaetifs Jun 2 '17 at 2:26

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