# Does the alternating series $\sum_{n=1}^{\infty}(-1)^n$ converge?

Does the series

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

converge? I was trying to use this as a convergent majorant for proving convergence of $$\sum_{n=1}^\infty (-1)^n \frac {\ln(n)} n$$ but I'm not sure if that would work. Any help is appreciated!

• It ($\sum\limits_{n=1}^{\infty}(-1)^n$) doesn't converge since $(-1)^n$ does not go to $0$. Remember that $\sum\limits_{n=1}^{\infty}a_n$ can only converge if $\lim\limits_{n\to\infty}a_n=0$. That series is an interesting series called Grandi's series though, and you can find more information about it at its Wikipedia page. – Minus One-Twelfth Apr 7 '19 at 11:59
• Suppose it did. What do you imagine it would convergent to? – MJD Apr 7 '19 at 11:59
• Second: You can prove convergence using a majorant, only if the proposed majorant is positive. – GEdgar Apr 7 '19 at 11:59
• @MJD I figured maybe 1 or $0$, but I'm not really convinced of it (hence my question here). – Benjamin Caris Apr 7 '19 at 12:02
• @KaviRamaMurthy Did not. However, I see what you mean now: it is alternating between -1 and 0, so indeed it does not converge. – Benjamin Caris Apr 7 '19 at 12:05

## 2 Answers

Think about what $$\sum_{n=1}^{\infty}(-1)^{n}$$ means

You get $$-1 + 1 -1 + 1...$$

Does that converge to a number? Clearly not, it goes back and forth between $$-1$$ and $$0$$

A convergent sum will get closer and closer to one particular value, and can in fact get arbitrarily close to that value, which clearly our sum does not.

Given $$n\geq 1$$, you have that $$\frac{\ln n}{n}\geq 0$$. Moreover, $$n\mapsto \frac{\ln n}{n}$$ is a decreasing function, which tends to $$0$$ as $$n\to\infty$$. Thus, you may apply the Alternating Series Test to conclude that the series $$\sum_{n=1}^{\infty}(-1)^{n}\frac{\ln n}{n}$$ converges.