# Alternating Series Convergence / Divergence

$$\sum_{n=1}^\infty (-1)^n$$ - I know this doesn't converge and I want to prove it. I am using the null sequence test as the limit as $$n$$ tends to $$\infty$$ of $$(-1)^n$$ doesn't equal 0 which I am trying to prove.

$$1.$$ First I thought I could let $$a_n := (-1)^n$$ and note that the two subsequences $$a_{ 2k}$$ and $$a_{2k-1}$$ have two distinct limits where $$k$$ is a natural number, them being $$1$$ and $$-1$$, so $$a_n$$ clearly diverges. Hence, if it diverges it doesn't converge to $$0$$, meaning the series diverges by null sequence test. $$\square$$

$$2.$$ Assume for sake of contradiction that $$a_n = (-1)^n$$ converges to a limit $$l$$ where $$l$$ is a real number. Let $$\epsilon = 1$$. Now, for all $$\epsilon > 0$$, there exists a natural number $$N$$ such that $$|(-1)^n-l|<1$$.

Note that when $$n$$ is even, $$(-1)^n = 1$$, so we have $$|1-l| < 1$$. When $$n$$ is odd, we have $$|-1-l| < 1$$. Now $$2 = |1-l + 1+l| <= |1-l| + |1+l| = |1-l| + |-1-l| < 1 + 1 = 2$$, so $$2<2$$ which is a contradiction. Hence, $$a_n$$ diverges so doesn't converge to $$0$$, meaning the series diverges by null sequence test. $$\square$$

• Convergence of the series is a convergence of partial sum, not element. Nov 18, 2022 at 11:43
• @openspace Okay, how is that relevant here ? For $2$, I know that the sequence tending to $0$ is necessary for series convergence, which is why I proved that the sequence does not converge to $0$
– user1071088
Nov 18, 2022 at 11:55
• Note that $a_n \to 0$ is equivalent to $|a_n| \to 0$. Nov 18, 2022 at 12:29
• @MartinR ohhhh yeah I remember that. That would make the question a lot easier. Because I didn’t notice that I’ll stick with my proof $1$ but I’ll keep what you said in mind for next time
– user1071088
Nov 18, 2022 at 12:35

Nikita, both of your ideas look good. Either one on its own is sufficient to show that the sequence diverges, and thus the series diverges. Just a couple little things on the writing:

1. In the first one, I would replace "Hence, if it diverges" with "Hence, since it diverges", since you just concluded in the previous sentence that it does in fact diverge.
2. In the second one, it's a bit confusing that you say "Let $$\epsilon = 1$$" and then continue with "Now, for all $$\epsilon > 0$$". Once you fix your $$\epsilon$$, you don't want to reassign it to something else. Your argument still makes perfect sense though if you just say "Let $$\epsilon = 1$$. Then there exists a natural number $$N$$ ..." If you think the person reading your proof will be really pedantic about it, then you could throw in "Let $$\epsilon = 1$$. Then, since $$\epsilon > 0$$, there exists a natural number $$N$$ ..."

Hope this helps!

EDIT: If we want to consider the problem directly rather than showing that the sequence $$a_n$$ doesn't converge (which is sufficient), then we can directly observe that the partial sums don't converge since for all $$N \in \mathbb{N}$$, we have $$\left|\left(\sum_{n=1}^{N+1} (-1)^n\right) - \left(\sum_{n=1}^N (-1)^n\right)\right| = |(-1)^{N+1}| = 1,$$ so the sequence of partial sums is not Cauchy, and therefore does not converge.

• $1$ Understood. $2$ Yeah that is sloppy I agree, I'll fix it. Thanks for your help, I feel like $1$ is better because it's simpler so I will use that argument.
– user1071088
Nov 18, 2022 at 11:54
• No problem Nikita! I agree that 1 is simple and really easy to understand. Nov 18, 2022 at 11:57
• If the sequence doesn't converge, it's incorrect to say it diverges right because you can only assume that for a series. So I would have to say that since $a_n$ doesn't converge to a real number that it doesn't converge to $0$....
– user1071088
Nov 18, 2022 at 14:59
• I'm not quite sure what you're getting at. One rule for series is that if $a_n \not\rightarrow 0$, then $\sum_{n=1}^\infty a_n$ is not well-defined. This is what I thought you were referring to when saying "null sequence test", but maybe I misunderstood Nov 18, 2022 at 22:09
• yeah perfect, that’s what I meant
– user1071088
Nov 18, 2022 at 23:15