Determine whether or not the series $\sum\limits_{n=2}^\infty (\frac{n+4}{n+8})^n$ converges or diverges 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: Write 
$$a_n=\left(\frac{n+4}{n+8}\right)^n=\frac{\left(1+\frac{4}{n}\right)^n}{\left(1+\frac{8}{n}\right)^n}\to \frac{e^4}{e^8}=e^{-4}\ne 0\implies \,\,\text{the series diverges}$$
A: You can rewrite the expression for a term in this series using long division:
$$\frac{n+4}{n+8}=1-\frac{4}{n+8}$$
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:
$$\begin{align}\lim_{n\to\infty}\left(\frac{n+4}{n+8}\right)^n &= \exp\ln\lim_{n\to\infty}\left(\frac{n+4}{n+8}\right)^n\\
&=\exp\lim_{n\to\infty}n\ln\left(\frac{n+4}{n+8}\right)\\
&=\exp\lim_{n\to\infty}\frac{\ln(n+4)-\ln(n+8)}{n^{-1}}\\
&=\exp\lim_{n\to\infty}\frac{\frac{1}{n+4}-\frac{1}{n+8}}{-n^{-2}}\\
&=\exp\lim_{n\to\infty}\left(\frac{n^2}{n+8}-\frac{n^2}{n+4}\right)\\
&=\exp\lim_{n\to\infty}\frac{-4n^2}{n^2+12n+32}\\
&=\exp(-4)
\end{align}$$
A: $\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
$.
