$\sum_{n=1}^{\infty} (-1)^n\cos (1/n^2)$ Converges or Diverges using divergence test.
$\sum_{n=1}^{\infty} (-1)^n\cos (1/n^2)$
For this sum, if I use the divergence test, am I allowed to say that:
$\lim_{x \to \infty} (-1)^n\cos (1/n^2)$= $\lim_{x \to \infty} (-1)^n*\lim_{x \to \infty}\cos (1/n^2)$
Which diverges because first limit doesn't exist and the second one is equal to 1?
This means $\lim_{x \to \infty} (-1)^n\cos (1/n^2) \neq 0$
So it diverges.
If I can't do that, what convergence test should I should?
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$\begingroup$It is true that the sequence of terms $(-1)^n \cos(1/n^2)$ does not converge to zero, and you may use that fact to conclude that the series diverges.
However, your proof that it does not converge to zero is faulty. The hypothesis of the theorem$$\lim_{n \to \infty} x_n y_n = \lim_{n \to \infty} x_n \cdot \lim_{n \to \infty} y_n $$is that both limits on the right hand side exist, but $(-1)^n$ has no limit. So you'll need to do something else there...
$\endgroup$ 2 $\begingroup$Alternative approach:
As $~\displaystyle n \to \infty, ~~\cos\left(\frac{1}{n^2}\right) \to 1$.
Therefore, the terms of the series approach
$+1, -1, +1, - 1, \cdots$.
If needed, $\epsilon, \delta$ can be called off the bench to complete the analysis, under the hypothetical conclusion that there exists some limit $L$ such that the series converges to $L$.
Edit
See also the comment that I left, just after the answer of Lee Mosher.
Another way to phrase your exact idea which is much more clear and I think no one will argue with is the following:
Suppose the limit
$$\lim_{n\to\infty} (-1)^n\cos\left(\frac{1}{n^2}\right)$$
does exist. Then we have that the following exists:
$$\frac{\lim_{n\to\infty}(-1)^n\cos\left(\frac{1}{n^2}\right)}{\lim_{n\to\infty}\cos\left(\frac{1}{n^2}\right)}=\lim_{n\to\infty}\frac{(-1)^n\cos\left(\frac{1}{n^2}\right)}{\cos\left(\frac{1}{n^2}\right)}=\lim_{n\to\infty}(-1)^n.$$
But we know already that the limit on the right hand side does not exists, so we have a contradiction, and the limit does not exist.
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