If a sequence has two subsequences that do not both converge to the same limit, then the sequence does not converge.
This can be proven using the $\epsilon-\delta$ definition of convergence:
- Let $a_n$ converge to $L$, and let $\{a_{n_k}\}_{k\to\infty}$ be a subsequence of $a_n$.
- Let $\epsilon > 0$.
- Then, because $a_n$ converges to $L$, there exists some $N$ such that if $n>N$, then $|a_n - L|<\epsilon$.
- Because $n_k$ is an increasing sequence of integers (by definition of a subsequence), there exists such $K$ that $n_K > N$.
- Then, if $k > K$, we have $n_k > n_K > N$, and from the previous point, we get that $|a_{n_k} - L|<\epsilon$, meaning that $a_{n_k}$ converges to $L$.
However, your case is special in that there is an overlap of sequences, for example $a_6$ is in the first and third sequence, and $a_9$ is in the second and third sequence. In fact, the third sequence alternates between the first two sequences, and you can use this to prove all three limits must be equal.
Yes, your proof is valid.
But
First, let's notice that $\color{red}{−1≤cos(πn)≤1}$, and that $cos(πn)$ oscillates between $−1$ and $1$
The first, in red, isn't necessarily relevant and you never used it. And it won't help you. The second, depending on the whim of the grader, may or may not need to be verified or more formally defined.
I'd say: $\cos(n\pi) = (-1)^n$ which equals $+1$ if $n$ is even, and equals $-1$ if $n$ is odd.
let's take the subsequence bn=(1,−1,1,−1,…)=(−1)n+1 of an, which also oscillates between −1 and 1.
This isn't invalid but I don't see why you are doing this. There is no reason you sequence starts on $1$ rather than $-1$. Just use $a_n$ and don't bother with this.
If we take two subsequences of bn, let them be b′n=1 and b′′n=−1
Okay, but it might be better to formally describe how to do this.
Let $b'_n = a_{2n}= \cos (2n\pi) = 1$ and let $b''_n = a_{2n+1} = \cos((2n+1)\pi) = -1$.
we will find .....
whoa... if you say "we will find" you're just asking for the grader to so "Oh, yeah. When will we find that?" :)
.... that these two subsequences do not converge to the same limit.
Just say that they do converge to different limits.
I'd be a real sadist if I required you to prove that but it's enough to say $b'_n$ converges to $1$ (because it is constant) and $b''_n$ converges to $-1$.
Since bn has two subsequences that do not converge to the same limit, bn is a divergent sequence.
You should cite the theorem that states this is so.
Since bn is divergent, and by the divergence criteria for sequences, an is then divergent.
Again there was no reason to have ever introduce the $b_n$.
Anyway... I'd give full marks but with the comments I just gave.
Best Answer
If you can partition a sequence into finitely many subsequences, each of which converges to $L$, then the original sequence must converge to $L$ as well. This is clear from the following definition of the limit: $a_n \rightarrow L$ iff for all $\epsilon>0$ there exists $N_\epsilon$ such that $|a_n - L| < \epsilon$ whenever $n \ge N_\epsilon$. But then for any $\epsilon > 0$, because the $i$-th subsequence converges to $L$, it is within $\epsilon$ of the limit for $n \ge N^{(i)}_\epsilon$; so the sequence itself is within $\epsilon$ of the limit for $n \ge N_\epsilon = \max_{i}N^{(i)}_\epsilon$.
However, the result does not hold for a partition into infinitely many subsequences. For instance, consider the case where $a_n=1$ when $n$ is prime and $a_n=0$ otherwise. This can be partitioned into infinitely many subsequences, where $a_n$ is in the $i$-th subsequence if the smallest prime factor of $n$ is the $i$-th prime. (Just put $a_1$ anywhere.) Each subsequence converges (immediately) to zero, but the original sequence does not converge, because it has sporadic $1$'s as far out as you care to look.