<s>[INST] limit $\lim_{n \to \infty} \frac{(2n+1)(2n+2) n^n}{(n+1)^{n+2}}$? Wolfram alpha says that this limit (arising from a ratio test to determine the radius of convergence of a series) should be $4/e$. How does it get this result?
 [/INST] Notice that
$$\frac{2n+1}{n+1} = 2 - \frac{1}{n+1}.$$
Then
\begin{eqnarray*}
\lim_{n \to \infty} \dfrac{(2n+1)(2n+2) n^n}{(n+1)(n+1)^{n+1}} &=& \lim_{n \to \infty} \left(2-\frac{1}{n+1}\right)\dfrac{2(n+1)n^n}{(n+1)^{n+1}} \\
&=& \lim_{n \to \infty} \left(2-\frac{1}{n+1}\right)\dfrac{2n^n/n^n}{(n+1)^{n}/n^n} \\
&=&\lim_{n \to \infty} \left(2-\frac{1}{n+1}\right)\dfrac{2}{(1+\frac1n)^n} \\
&=& \frac{4}{e}
\end{eqnarray*}
</s>