Solutions - 0180-06

For very large $n$, we expect the exponentials to dominate, and the series looks like ${\displaystyle \frac{e^n}{e^{2n}}}={\displaystyle \frac{1}{e^n}}$. This will yield a converging geometric series. Anyway, let us choose $b_n={\displaystyle \frac{1}{e^n}}$ as the comparison series.

$$a_n\cdot b_n^{-1}=\frac{e^n+n}{e^{2n}-n^2}\cdot \frac{e^n}{1}\gg \gg \frac{e^{2n}+n{e}^n}{e^{2n}-n^2}$$

Now divide above and below by the leading power:

$$\begin{array}{c}\frac{e^{2n}+n{e}^n}{e^{2n}-n^2}\gg \gg \frac{e^{-2n}}{e^{-2n}}\cdot \frac{e^{2n}+n{e}^n}{e^{2n}-n^2}\\ \\ \gg \gg \frac{1+n{e}^{-n}}{1-n^2{e}^{-2n}}\end{array}$$

By L’Hopital’s Rule, we find that:

$$\begin{array}{c}n{e}^{-n}\stackrel{n\to \infty }{⟶}0\\ \\ n^2{e}^{-2n}\stackrel{n\to \infty }{⟶}0\end{array}$$

Therefore:

$$\frac{1+n{e}^{-n}}{1-n^2{e}^{-2n}}\stackrel{n\to \infty }{⟶}1=L$$

Since $0<L<\infty$, the LCT says that both series converge, or both diverge.

Now ${\displaystyle \sum \frac{1}{e^n}=\sum {\left(\frac{1}{e}\right)}^n}$ and this is geometric with $r=\frac{1}{e}<1$. Therefore it converges, and the original series must converge too.