Question 88
Prove that
$$
\lim_{n \rightarrow +\infty} \frac{\ln \left(1 + 2e + 4e^{4} + 6e^{9} + \ldots + 2n e^{n^{2}}\right)}{n^{2}} = 1.
$$
$$
\lim_{n \rightarrow +\infty} \frac{\ln \left(1 + 2e + 4e^{4} + 6e^{9} + \ldots + 2n e^{n^{2}}\right)}{n^{2}} = 1.
$$