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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?

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    Try using the Bernoulli inequality. What do you obtain? – 2017-02-22

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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 $\exp$, $$ \left(1 + \frac{1}{n+2}\right)^{3n\log(n)} \xrightarrow[n\to\infty]{} \infty. $$

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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.

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    That's a good intuition, but the argument needs to be much more formal and rigorous than that. Here, you are basically saying "if $u_n\to \ell$ and $v_n\to \ell'$, then $u_n^{v_n}\to \ell^{\ell'}$. This is **false** in general: exactly, take $u_n = 1+\frac{1}{n}$ and $v_n=n$. – 2017-02-22
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    @ClementC, I agree that this is not rigorous, but in your counterexample "$1^\infty$" is an "indeterminate form" whereas the "$e^\infty$" form of the limit expressed in this solution is not. – 2017-02-23
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    I know this cas3 can actually be made rigorous - I merely wanted to point out that these intuitions can be really misleading. – 2017-02-23