Alternating Series Convergence / Divergence

Question

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

Answer 1

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.