Does the series converge or diverge?

The series $\sum\limits_{n=2}^\infty (\frac{n+4}{n+8})^n$

I tried the root test to get rid of the nth power but the limit equaled $1$ so the test is inconclusive. How else can I determine if this series converges or diverges?



$$a_n=\left(\frac{n+4}{n+8}\right)^n=\frac{\left(1+\frac{4}{n}\right)^n}{\left(1+\frac{8}{n}\right)^n}\to \frac{e^4}{e^8}=e^{-4}\ne 0\implies \,\,\text{the series diverges}$$

  • $\begingroup$ This is much prettier than my answer. Nice one. +1 $\endgroup$ – G Tony Jacobs Jul 7 '15 at 3:51
  • $\begingroup$ @GTonyJacobs Wow, thank you! I'm 6 days out of surgery and recovering at home. That just made my day. $\endgroup$ – Mark Viola Jul 7 '15 at 3:53

You can rewrite the expression for a term in this series using long division:


We recognize the limit:

$$\begin{align}\lim_{n\to\infty} \left(1-\frac{4}{n+8}\right)^n &= \lim_{n\to\infty} \left(\left(1-\frac{4}{n+8}\right)^{n+8}\right)^{\frac{n}{n+8}}\\ &=\left(e^{-4}\right)^1\\ &\neq 0 \end{align}$$

Therefore, by the divergence test, this series diverges.

Alternatively, you could just use L'Hopital's rule directly to calculate the same limit:

$$\begin{align}\lim_{n\to\infty}\left(\frac{n+4}{n+8}\right)^n &= \exp\ln\lim_{n\to\infty}\left(\frac{n+4}{n+8}\right)^n\\ &=\exp\lim_{n\to\infty}n\ln\left(\frac{n+4}{n+8}\right)\\ &=\exp\lim_{n\to\infty}\frac{\ln(n+4)-\ln(n+8)}{n^{-1}}\\ &=\exp\lim_{n\to\infty}\frac{\frac{1}{n+4}-\frac{1}{n+8}}{-n^{-2}}\\ &=\exp\lim_{n\to\infty}\left(\frac{n^2}{n+8}-\frac{n^2}{n+4}\right)\\ &=\exp\lim_{n\to\infty}\frac{-4n^2}{n^2+12n+32}\\ &=\exp(-4) \end{align}$$

  • $\begingroup$ Are you allowed to ignore the $\frac{n}{n+8}$ because it tends to 1? $\endgroup$ – user217285 Jul 7 '15 at 3:34
  • 1
    $\begingroup$ Yeah, more or less. I'm not ignoring it, so much as using the rule that $\lim f(x)^{g(x)}=(\lim f(x))^{\lim g(x)}$, in cases where both limits are defined, and where we don't get an indeterminate form. $\endgroup$ – G Tony Jacobs Jul 7 '15 at 3:37

$\begin{array}\\ a_n &=\left(\dfrac{n+u}{n+v}\right)^n\\ &=\left(\dfrac{n+v-v+u}{n+v}\right)^n\\ &=\left(1+\dfrac{-v+u}{n+v}\right)^n\\ &\to e^{u-v}\\ \end{array} $

since $(1+\frac{a}{x})^x \to e^a $.


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