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For example, in this problem (trying to determine convergence):

$$\sum\limits_{n=1}^\infty \frac{\cos\left(\dfrac{1}{n}\right)}{n^2}$$

Can I just compare this ($a_n$):

$$\frac{\cos\left(\dfrac{1}{n}\right)}{n^2}$$

to this? ($b_n$):

$$\cos\left(\dfrac{1}{n}\right)$$

$\cos(1/n)$ converges at 1 and $a_n \le b_n$ so the original summation converges, yes? I was just wondering the rules for choosing $b_n$. Can I just choose something completely arbitrary and not at all related to the original equation?

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You have to chose $b_n$ so that the series $\sum b_n$ converges. In your example, since the sequence $\cos(\frac{1}{n})$ converges to $1$, the series $\sum \cos(\frac{1}{n})$ does not converge. But you can show $|a_n| \le \frac{1}{n^2}$ and the series $\sum \frac{1}{n^2}$ converges. –  Joel Cohen Nov 8 '11 at 6:35
    
It only has to be related enough to the original summand to do what we need. If we think $\sum a_n $ converges, then we need $b_n$ to satisfy $ a_n \leq b_n $ and also $ \sum b_n < \infty . $ Other than those, it can be arbitrary. –  Ragib Zaman Nov 8 '11 at 6:36

2 Answers 2

up vote 3 down vote accepted

Note that $|\cos(1/n)|\le 1$. So by comparison with $\sum_1^\infty \frac{1}{n^2}$, our series converges absolutely, and hence converges.

Of course we need not have taken absolute values, $\cos(1/n)$ is always positive. But for example exactly the same argument would work for $\sum_1^\infty\frac{\cos n}{n^2}$, and $\cos n$ varies quite wildly in the interval $(-1,1)$.

It is hard to give general rules, but $\frac{1}{n^2}$ goes down reasonably fast, so if $\cos(1/n)$ doesn't spoil things, we will be OK. And $\cos(1/n)$ cannot spoil things, since it never gets very big.

Comparison with the series $\sum \cos(1/n)$ is no good, since it diverges. The terms don't even approach $0$, they approach $1$, so you are summing a whole bunch of things that are near $1$. It is clear that the sum blows up.

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It's true that $$\frac{\cos(1/n)}{n^2} \leq \cos(1/n)$$ but the series $$\sum_{n=1}^{\infty}\cos(1/n)$$ does not converge so you cannot use the comparison test in this case. Often, if you are trying to prove that a sum of the form $$\sum_{n=1}^{\infty} a_n$$ converges, you can successfully bound it from above by a geometric series or $p$-series with $p > 1$. In this case you want $$ a_n \leq c r^n\quad\text{or}\quad a_n \leq \frac{c}{n^p}$$ for all $n\ge 1$ and for some constants $c,r$ with $|r| < 1$ or $p>1$.

If you want to show that the sum diverges, you can often bound it below by the harmonic series. In that case you want $$ a_n \ge \frac{c}{n}$$ for all $n\ge 1$ for some constant $c$.

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