Bolzano–Weierstrass theorem When proving that a certain sequence does not converge, is it enough to show that there exist 2 sub-sequences, one that converges to a limit $L$, and the other diverging (increases without bound for example) to prove this, or must I show that there exist two different sub-limits?
(I'm using the rule that every sub-sequence of a convergent sequence converges to the same limit.)
For example:
When proving that $$a_n = \cos\left(\frac{\pi n^2} {2n+3}\right)$$ does not converge, I found two sub-sequences, $b_n$ and $c_n$, where $b_k = \frac1 k$ and $c_k = \sqrt{k}$, in which $a_{b_k}$ converges to $0$ and $a_{c_k}$ diverges. 
EDIT
Just realized the above aren't indices at all... since $k$ must be integers. I've tried subs-sequences of even and odd integer indices for $k$ but don't see a pattern for the function. 
 A: Hint: Note that
$$\frac{n^2}{2n+3}=\frac{n^2+3n/2}{2n+3}-\frac{3n/2}{2n+3}=\frac{n}{2}-\frac{3n/2}{2n+3}.$$
Let $n$ be large.  The second term has limit $3/4$ as $n\to \infty$.  Now multiply by $\pi$. Now let us explore the values of the cosine.  We are subtracting something very close to $3\pi/4$ from $n\pi/2$. So our angle has the same cosine as $\pi(n-4)/2$ plus something that approaches $\pi/4$.  
Look  at what happens when $n$ is a multiple of $4$. Then our cosine is very close to the cosine of $\pi/4$, which is $\sqrt{2}/2$. When $n$ is $2$ more than a multiple of $4$, then our cosine is very close to the cosine of $3\pi/4$, so it is near $-\sqrt{2}$. Thus we can use as subsequences the integers $n$ of the form $4k$, and the integers $n$ of the form $4k+2$. (There is a similar alternation with the odd $n$, but we don't need to look at them, since we already have non-convergence.) 
Remark Informally, $n^2/(2n+3)$ is "about" $n/2$ when $n$ is large, and $\cos(n\pi/2)$ is quite different when $n$ is even than when $n$ is odd. However, "about" is much too vague for our purposes. 
Even though it is not necessary for the argument, we can go further in our decomposition, by noting that 
$$\frac{(3n/2)}{2n+3}=\frac{3}{4}-\frac{9/4}{2n+3}.$$ 
The best approach to our first decomposition is not the "magic" adding and subtracting of $3n/2$. Instead, divide the polynomial $x^2$ by the polynomial $2x+3$ using the ordinary "long division" process. 
Alternately, let $m=2n+3$. Then $n=(m-3)/2$, and therefore $n^2=(1/4)(m-3)^2$. Expand. Dividing by $m$ is easy.  
A: Yeah, remember that if $(S_n)$ is a convergent sequence, then each of its subsequences converges as well. So, if you can find just a single divergent subsequence, then you're done. Now,like you said, if you are able to find two subsequences that converge to two distint limits however, you are also done as the limit of any sequence is unique. 
