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 $e$

$$a_n=\left(1+\frac{1}{n}\right)^{n^2}\implies \log(a_n)=n^2 \log\left(1+\frac{1}{n}\right)$$ 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
$$\log(a_{n+1})-\log(a_n)=1-\frac{1}{3 n^2}+O\left(\frac{1}{n^3}\right)$$
$$\frac{a_{n+1} }{a_n }=e^{\log(a_{n+1})-\log(a_n)}=e-\frac{e}{3 n^2}+O\left(\frac{1}{n^3}\right)$$
