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How to test the convergence of $\sum_{n=1}^\infty \dfrac{n\log n}{e^n}?$

Sriti Mallick
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5 Answers5

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As usual, I love D'Alemberts ratio test:

$$a_n:=\frac{n\log n}{e^n}\Longrightarrow \frac{a_{n+1}}{a_n}=\frac{(n+1)\log(n+1)}{e^{n+1}}\frac{e^n}{n\log n}=$$ $$=\frac{n+1}{n}\frac{\log(n+1)}{\log n}\frac{1}{e}\xrightarrow[n\to\infty]{}\frac{1}{e}<1$$

and thus the series converges.

DonAntonio
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    Does $\dfrac{\log(n+1)}{\log n}\to 1?$ How can it be proved? – Sriti Mallick Mar 10 '13 at 17:26
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    For example with L'Hospital, say: $$\lim_{x\to\infty}\frac{\log(x+1)}{\log x}=\lim_{x\to\infty}\frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x\to\infty}\frac{x}{x+1}=1$$ – DonAntonio Mar 10 '13 at 17:30
  • Or by the squeeze theorem, too, using monotony of $\log n,$: $$1=\frac{\log n}{\log n}\le\frac{\log(n+1)}{\log n}\le\frac{\log 2n}{\log n}=\frac{\log 2}{\log n}+\frac{\log n}{\log n}\xrightarrow[n\to\infty]{} 1$$ – DonAntonio Mar 10 '13 at 17:36
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You can easily show that $$ \frac{n\log n}{e^n}=O\left(\frac{1}{n^2}\right). $$ All you have to do is check that $$ \lim_{n\rightarrow +\infty}n^2 \cdot\frac{n\log n}{e^n}=0. $$

Then conclude the series converges (absolutely, of course) by comparison with the Riemann series $\sum_{n\geq 1} \frac{1}{n^2}$.

Julien
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    you confused me with your edit, maybe cause i don't like the $=\mathcal{O}()$ notation, for a freshan wouldn't it be easiert to show $n \log(n) <n^2$ and $e^n > n^4$ ? – Dominic Michaelis Mar 10 '13 at 17:15
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    @DominicMichaelis Yes, sure. But in many instances (not here, I agree) the big O notation is very useful and it makes it a bummer to avoid it. So the sooner you encounter it, the better, imo. – Julien Mar 10 '13 at 17:19
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    @DominicMichaelis I've indeed noticed the general preference here to manipulate inequalities when solving this kind of questions. I do it sometimes. But I find it artificial. So I don't always do it. Because I am not sure this is the best way to prepare the freshman to treat more delicate cases where easy inequalities don't suffice. By the way $e^n\geq n^4/4!$ is easy, $e^n\geq n^4$ is less easy. – Julien Mar 10 '13 at 17:27
  • +1 for the elegant approach. I think this question has got enough points, form yours to Don's. – Mikasa Mar 10 '13 at 17:36
  • @BabakS. Thanks, Babak. – Julien Mar 10 '13 at 17:45
  • @DominicMichaelis By the way, aren't you german like Landau? You should promote your heritage! – Julien Mar 10 '13 at 18:09
  • @julien j'aime ton carré rouge! :) – Bruno Joyal Mar 10 '13 at 18:37
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    @Bruno Ah! Someone understood the reference! This is indeed more or less a private joke. – Julien Mar 10 '13 at 18:40
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Another Hint:

$$\lim_{n\to\infty}n^{2}\times\frac{n\log(n)}{\exp(n)}=0<\infty$$ so it converges.

Mikasa
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First note that $e^n > n^4$ for $n$ large enough. That is, first prove that there is a $K$ such that $e^n > n^4$ for all $n\geq K$.

Then you have for $n\geq K$ also that $\log(n) < n$ and so $$ \frac{n\log(n)}{e^n} < \frac{n\log(n)}{n^4} < \frac{n^2}{n^4} = \frac{1}{n^2}. $$ Hence you can compare the sequence to $\sum\frac{1}{n^2}$.

Thomas
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The ratio test is probably the best tool to use in this case (and in many other examples), but you may also use the integral test.

Let $f(x)=x\log x/e^x$. Clearly $f(x)$ is positive for all $x > 1$. Moreover $f(x)$ is decreasing at least on $[e,\infty)$. Indeed, $$f'(x) = \frac{1+(1-x)\log(x)}{e^x}$$ so if $x\geq e$, then the numerator of $f'(x)$ is less than $1+(1-e)<0$. Hence $f'(x)<0$ for $x\geq e$, and $f$ is decreasing. Hence, we may apply the integral test on the series $\sum_{n=1}^\infty f(n)$.

The series $\sum_{n=1}^\infty f(n)$ converges if and only if $\sum_{n=3}^\infty f(n)$ converges and this series, by the integral test, converges if and only if $\int_3^\infty f(x)dx$ converges. Now:

$$\int_3^\infty f(x) dx = \lim_{N\to \infty} \int_3^N \frac{x\log x}{e^x} dx \leq \lim_{N\to \infty} \int_3^N \frac{x^2}{e^x} dx = \lim_{N\to\infty} [-e^{-x}(x^2+2x+2)]_3^N$$ $$ = \lim_{N\to\infty} -\frac{N^2+2N+2}{e^N}+\frac{17}{e^3}=\frac{17}{e^3}.$$ Since the improper integral converges, we conclude that the series $\sum_{n=1}^\infty f(n)$ converges as well.

Álvaro Lozano-Robledo
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