Using the ratio test on $\sum_{n=1}^{\infty} \left(1+\frac{1}{n}\right)^{n^2}$ I already know that I should do the root test on this series due to the $n$th power, but I want to see if I can establish the result using the ratio test first. (Or would I always be stuck with only one kind of test)
I get something like this: $$\lim_{n \to \infty} \left|\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}\right|$$
But then I'm not sure how (or even if I could) simplify this. I've tried on Wolfram Alpha to make it simplify but the best I can get is numerical approximations that seem to converge to $e$
 A: To find the limit $$ \begin{align}\displaystyle  \dfrac{\left(1+\dfrac{1}{(n+1)}\right)^{(n+1)^2}}{\left(1+\dfrac{1}{n}\right)^{n^{2}}} \cdot \dfrac{\left( 1 + \dfrac{1}{n} \right)^{(n+1)^2}}{\left( 1 + \dfrac{1}{n} \right)^{(n+1)^2}} &= \left( \dfrac{n(n+2)}{(n+1)^2} \right)^{(n+1)^2} \cdot \left( 1 + \dfrac{1}{n} \right)^{2n+1} \\
&= \left( 1 - \dfrac{1}{(n+1)^2} \right)^{(n+1)^2} \cdot \left( 1 + \dfrac{1}{n} \right)^{2n+1} \end{align} $$
A: First, note that
$$
\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}
= 1-\frac{1}{(n+1)^2} \tag{1}
$$
We will use that to make the "right" exponent appear, since we want to use the known limit $\lim_{m\to\infty} (1+\frac{u}{m})^m = e^u$.
Now, since $n^2=(n+1)^2-(2n+1)$, we can rewrite
$$\begin{align*}
\frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}
&= \left(1+\frac{1}{n}\right)^{2n+1}\cdot \frac{\left(1+\frac{1}{n+1}\right)^{(n+1)^2}}{\left(1+\frac{1}{n}\right)^{(n+1)^2}}
= \left(1+\frac{1}{n}\right)^{2n+1} \left(\frac{1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n^2} \\
&= \color{red}{\left(1+\frac{1}{n}\right)}\cdot \color{blue}{\left(1+\frac{2}{2n}\right)^{2n}}\cdot \color{green}{\left(1-\frac{1}{(n+1)^2}\right)^{(n+1)^2}} \tag{2}
\end{align*}$$
This is great! We know that
$$
\lim_{n\to\infty} \left(1+\frac{1}{n}\right) = \color{red}{1} \tag{3}
$$
and
$$
\lim_{n\to\infty} \left(1+\frac{2}{2n}\right)^{2n} = \color{blue}{e^{2}} \tag{4}
$$
and
$$
\lim_{n\to\infty} \left(1-\frac{1}{(n+1)^2}\right)^{(n+1)^2} = \color{green}{e^{-1}} \tag{5}
$$
so, combining (3), (4), and (5) into (2), we get
$$
\lim_{n\to\infty} \frac{\left(1+\frac{1}{n+1}\right)^{\left(n+1\right)^2}}{\left(1+\frac{1}{n}\right)^{n^2}}
= \color{red}{1} \cdot \color{blue}{e^{2}} \cdot \color{green}{e^{-1}} = \boxed{e}
$$
A: $$a_n=\left(1+\frac{1}{n}\right)^{n^2}\implies \log(a_n)=n^2 \log\left(1+\frac{1}{n}\right)$$ Now, by Taylor
$$\log(a_n)=n-\frac{1}{2}+\frac{1}{3 n}-\frac{1}{4 n^2}+O\left(\frac{1}{n^3}\right)$$ Apply it a second time and continue with Taylor
$$\log(a_{n+1})-\log(a_n)=1-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} }{a_n }=e^{\log(a_{n+1})-\log(a_n)}=e-\frac{e}{3 n^2}+O\left(\frac{1}{n^3}\right)$$
