Have a question that I'm stuck on here.

Let $$r_n= \frac{\sqrt{n}}{n!}\left(\frac{n}{e}\right)^n$$

Express $\log\left(r_{n+1}/r_n\right)$ as simply as possible. For this I got $\left(\frac{1}{2}+n\right)\log\left(1+\frac{1}{n}\right)-1$.

b) Verify that the following limit exists and calculate it:

$$\lim_{x\rightarrow 0}\frac{\left(1+\frac{x}{2}\right)\log(1+x)-x}{x^3}.$$ I got $\dfrac{1}{12}$, which I'm pretty sure is correct. I used L'Hôpital's rule.

c) Use your answers to parts a) and b), and the comparison test to show that $\displaystyle\sum_{n=1}^\infty \log\left(\frac{r_{n+1}}{r_n}\right)$ converges. Deduce that $\displaystyle\lim_{n\rightarrow\infty} r_n$ exists.

Struggling on part c, can't explicitly link it to parts a and b, did not find a suitable series for the comparison test.

Sorry for lack of LaTex, if someone could edit that would be great. I tried to use LaTex on the first question but it screwed up.

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    $\begingroup$ If you click on edit, you'll see what has been done by way of TeX. Mostly, math has to be inside dollar signs, log has to be \log, and r_n+1 has to be r_{n+1}. $\endgroup$ Mar 10, 2013 at 22:30
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    $\begingroup$ You can also see the $\TeX$ source by right-clicking on a formula and selecting "Show Math As:TeX Commands". $\endgroup$
    – joriki
    Mar 10, 2013 at 22:35
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    $\begingroup$ Is it $\log\left(\dfrac{r_{n+1}}{r_n}\right)$ (as edited by Gerry Myerson) or $\log\left(r_n+\dfrac{1}{r_n}\right)$ (as per last edit by Cortizol)? $\endgroup$ Mar 10, 2013 at 22:53
  • $\begingroup$ @Américo: I reinstated the former, which was clearly the intended meaning; the OP was merely missing the grouping braces around the index. $\endgroup$
    – joriki
    Mar 11, 2013 at 4:11

1 Answer 1


I think the only thing you're missing is that if you take the expression in b) and substitute $x=1/n$ so it becomes a limit as $n\to\infty$, it becomes a limit very closely related to what you have in a).


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