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

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 $n\to \infty$

• Note that $2^n/n^2\to+\infty$ – Robert Z May 14 '18 at 17:43
• @RobertZ Means it is divergent. Q1. How to check $2^n/n^2\to+\infty$, Q2. Is it sufficient that $2^n/n^2\to+\infty$ implies series is divergent? – user1942348 May 14 '18 at 17:54
• But D' Alembert's theorem is applicable for the series of positive terms? Here it is an alternating series. – user1942348 May 14 '18 at 17:58
• Sorry. I mean $\frac{|u_{n+1}|}{|u_n|}\to 2>1$ implies that $|u_n|\to +\infty$ and therefore the series is not convergent (a necessary condition for convergence is $u_n\to 0$). – Robert Z May 14 '18 at 18:03
• @user1942348 If the general terms of a series do not approach $0$, then the series diverges. Inasmuch as $\frac{(-1)^{n-1}2^n}{n^2}$ does not approach $0$, the series diverges. That is all we need. – Mark Viola May 14 '18 at 18:25

## 2 Answers

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.

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