Use the alternating series test to study the convergence of the following series sum_{n=1}^infty(-1)^{n+1}ne^{-n}

Question
Series
Use the alternating series test to study the convergence of the following series
$$\sum_{n=1}^\infty(-1)^{n+1}ne^{-n}$$

2021-02-03
Alternating series:
A series of the form
$$\sum_{n=1}^\infty(-1)^{n+1}na_n=a_0-a_1+a_2-a_3+...$$
where either all $$a_n$$ are positive or all $$a_n$$ are negative, is called an alternating series.
The alternating series test then says: if $$|a_n|$$ decreases monotonically and $$\lim_{n\rightarrow\infty}a_n=0$$ then the alternating series converges.
Moreover, let L denote the sum of the series, then the partial sum
$$S_k=\sum_{n=0}^k(-1)^na_n$$
approximates L with error bounded by the next omitted term:
$$|S_k-L|\leq|S_k-S_{k+1}|=a_{k+1}$$
i) $$a_n=ne^{-n}$$
$$=\frac{n}{e^n}>0$$
ii) $$n\to\infty\frac{n}{e^n}$$
$$n\to\infty\frac{1}{e^n}$$
$$=\frac{1}{\infty}$$
$$=0$$
iii)

Relevant Questions

Use the alternating series test to determine the convergence of the series
$$\sum_{n=1}^\infty(-1)^n\sin^2n$$
Consider the following series: $$\sum_{n=1}^\infty(-1)^n\frac{1}{n}$$
Use the Alternating Series Test to show this series converges.
Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series.
$$\sum_{n=2}^\infty\frac{(-1)^nn}{n^2-3}$$
Use the Alternating Series Test, if applicable, to determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty\frac{(-1)^n}{n^5}$$
Use the Root Test to determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty(\frac{3n+2}{n+3})^n$$
Use the Root Test to determine the convergence or divergence of the series.
$$\sum_{n=1}^\infty(\frac{n}{500})^n$$
Determine whether the series $$\sum a_n$$ an converges or diverges: Use the Alternating Series Test.
$$\sum_{n=2}^\infty(-1)^n\frac{n}{\ln(n)}$$
Use a power series to test the series $$\sum_{n=1}^\infty n^{\log x}$$ for convergence, where $$\sum$$ means summation
$$\sum_{n=1}^\infty\frac{5}{n+\sqrt{n^2+4}}$$
$$\sum_{n=1}^\infty\frac{n^{k-1}}{n^k+1},k>2$$