Determine if a sequence $\left(1 + \frac{1}{n+2}\right)^{3n\log(n)}$ converges or diverges, and if it converges find its limit? I've been given a sequence $\left(1 + \frac{1}{n+2}\right)^{3n\log(n)}$; aside from plugging in values into a calculator, I'm completely stuck on how to formally determine if this sequence converges or diverges, and to find its limit if it has one. Is there an intuitive approach to determining whether a sequence converges or diverges?
 A: This type of sequences with general term of the form $f(n)^{g(n)}$ can look confusing, because of the dependence on $n$ on both the base and the exponent. What follows is a detailed solution; it may not be the fastest in this particular case (there are "tricks" specific to this particular family of sequences that can yield a more succinct answer); but it's a pretty systematic approach.
To handle them, my advice is to systematically revert to the "exponential form":
$$
\left(1 + \frac{1}{n+2}\right)^{3n\log(n)}
= \exp\left( 3n\log(n) \log\left(1 + \frac{1}{n+2}\right)\right)
$$
Now, since $\exp$ is continuous, it will be sufficient to find the limit of $3n\log(n) \log\left(1 + \frac{1}{n+2}\right)$ when $n\to\infty$. This can be done much more easily:
$$\begin{align}
3n\log(n) \log\left(1 + \frac{1}{n+2}\right)
&= 3n\log(n) \cdot \frac{1}{n+2}\cdot \frac{\log\left(1 + \frac{1}{n+2}\right)}{\frac{1}{n+2}}
\\&= \underbrace{3\log n}_{(1)} \cdot \underbrace{\frac{n}{n+2}}_{(2)}\cdot \underbrace{\frac{\log\left(1 + \frac{1}{n+2}\right)}{\frac{1}{n+2}}}_{(3)}
\end{align}$$
We have three factors: when $n\to\infty$, $(1)$ goes to $\infty$; (2) goes to $1$; and finally, (3) goes to $1$, recalling that $\frac{\ln(1+x)}{x}\xrightarrow[x\to 0]{} 1$ (e.g., as definition of a derivative).
Thus,
$$
3n\log(n) \log\left(1 + \frac{1}{n+2}\right) \xrightarrow[n\to\infty]{} \infty
$$
and by continuity of $\exp$,
$$
\left(1 + \frac{1}{n+2}\right)^{3n\log(n)} \xrightarrow[n\to\infty]{} \infty.
$$
A: The usual process in limits of the kind $(1+a_n)^{b_n}$ where $a_n>0$, $a_n \to 0$ and $ b_n \to \infty$ is to look at the sequence with a general term $ b_n \cdot a_n$. This is because $$(1+a_n)^{b_n}= \left( (1+a_n)^{1/a_n} \right) ^{b_n \cdot a_n}.$$ Now remind yourself, what is the limit of $ (1+a_n)^{1/a_n}$? 
You can also see my comment for another similar approach.
