Investigate the convergence of the series $\sum_{n=1}^\infty (-1)^{n-1} \frac{2^n}{n^2}$

Question

Investigate the convergence of the two series

1. $$\sum_{n=1}^\infty \frac{2}{n^n}$$
2. $$\sum_{n=1}^\infty (-1)^{n-1} \frac{2^n}{n^2}$$

Attempt

1. $$\frac{u_{n+1}}{u_n}=\frac{n^n}{(n+1)^{n+1}}=\frac{1}{n(1+1/n)^{n+1}}\to0<1$$ then by D' Alemberts' test the series is convergent. Correct?

2. Let $v_n= \frac{2^n}{n^2}$ I want to check the convergence by Leibnitz's test. How to show that $\{v_n\}$ is monotonic decreasing and $v_n\to 0$ as $n\to \infty$

Answer 1

Your answer to (1) is correct.

For (2), the ratio is $\dfrac{\frac{2^{n+1}}{(n+1)^2}}{\frac{2^{n}}{n^2}} =\dfrac{2}{(1+1/n)^2} \to 2 $ so this sum diverges.

Answer 2

Note that for $v_n= \frac{2^n}{n^2}$, both conditions do not hold: $$1) \ v_{n+1}>v_n \iff \frac{2^{n+1}}{(n+1)^2}>\frac{2^n}{n^2} \iff 2n^2>(n+1)^2 \iff n^2>2n+1,n>2\\ 2) \ \lim_{n\to\infty} v_n=\lim_{n\to\infty} \frac{2^n}{n^2}\overbrace{=}^{L'H}\lim_{n\to\infty} \frac{2^n\ln 2}{2n}\overbrace{=}^{L'H} \lim_{n\to\infty} \frac{2^n\ln^22}{2}=\infty.$$