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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.


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.

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Actually one diverging subsequence is enough:) – savick01 Feb 25 '12 at 17:51
The Bolzano-Weierstrass theorem states that each bounded sequence has a convergent subsequence. It follows that an unbounded sequence has at least one diverging subsequence. – Josué Molina Feb 25 '12 at 17:53
The cosines stay between $-1$ and $1$, so you could not have found one of the subsequences you describe. And if $1/k$ describes indices, they are not going to infinity, indeed almost all are non-integers. – André Nicolas Feb 25 '12 at 17:57
@JosuéMolina: How does that follow? Sure, your conclusion is true, but I don't see that it follows. – Harald Hanche-Olsen Feb 25 '12 at 17:58
i tried setting each subsequence to even and odd integers but I don't see any pattern...the values jump all over the place – nofe Feb 25 '12 at 18:16
up vote 2 down vote accepted

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.

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is it usually a safe bet to first try the even/odd indices when looking for different sub-limits? – nofe Feb 25 '12 at 18:06
@nofe: My first hint was too vague. The behaviour of the trig functions is relevant. When there is bad behaviour, it is usually of the blowing up kind. But bouncing around alternately is probably the next most common kind. – André Nicolas Feb 25 '12 at 18:21
@nofe: There are all kinds of bad behaviour, such as wild oscillations. Approaching infinity (blowing up) means that our usual tools for analysis often break down. – André Nicolas Feb 25 '12 at 18:32
so how can I show this sequence diverges if the "about" statements are too vague? – nofe Feb 25 '12 at 18:35
@nofe: Recall that when $x$ is very close to $0$, $(\sin x)/x$ is close to $1$. So the ratio of $2^n\sin(1/n)$ to $2^n/n$ is close to $1$. However, you know that $2^n$ goes to infinity much faster than $n$. Thus the sequence you asked about diverges. Some would say instead it has limit $\infty$. – André Nicolas Feb 25 '12 at 19:34

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.

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