# How to prove that the series is converge?

## $\sum \frac{\cos \left(\frac{1}{k}\right)}{9 {k}^{2}}$

Apr 5, 2018

Converges by the Direct Comparison Test.

#### Explanation:

We can use the Direct Comparison Test, so far as we have

${\sum}_{n = 1}^{\infty} \cos \frac{\frac{1}{k}}{9 {k}^{2}}$, IE, the series starts at one.

To use the Direct Comparison Test, we have to prove that ${a}_{k} = \cos \frac{\frac{1}{k}}{9 {k}^{2}}$ is positive on $\left[1 , \infty\right)$.

First, note that on the interval $\left[1 , \infty\right) , \cos \left(\frac{1}{k}\right)$ is positive. For values of $x < \frac{\pi}{2} ,$ $\cos x$ is in the first quadrant (and therefore positive). Well, for $k \ge 1 , \frac{1}{k} < \frac{\pi}{2} ,$ so, $\cos \left(\frac{1}{k}\right)$ is indeed positive.

Furthermore, we can say $\cos \left(\frac{1}{k}\right) \le 1$, as ${\lim}_{k \to \infty} \cos \left(\frac{1}{k}\right) = \cos \left(0\right) = 1$.

Then, we can define a new sequence

${b}_{k} = \frac{1}{9 {k}^{2}} \ge {a}_{k}$ for all $k .$

Well,

${\sum}_{k = 1}^{\infty} \frac{1}{9 {k}^{2}} = \frac{1}{9} {\sum}_{k = 1}^{\infty} \frac{1}{k} ^ 2$

We know this converges by the $p -$series test, it is in the form $\sum \frac{1}{k} ^ p$ where $p = 2 > 1$.

Then, since the larger series converges, so must the smaller series.

Apr 5, 2018

It converges by the direct comparison test (see below for details).

#### Explanation:

Recognize that the range of cosine is [-1,1]. Check out the graph of $\cos \left(\frac{1}{x}\right)$:

graph{cos(1/x) [-10, 10, -5, 5]}

As you can see, the maximum value this will achieve will be 1. Since we're just trying to prove convergence here, let's set the numerator to 1, leaving:

$\sum \frac{1}{9 {k}^{2}}$

Now, this becomes a very simple direct comparison test problem. Recall what the direct comparison test does:

Consider an arbitrary series ${a}_{n}$ (we don't know if it converges/diverges), and a series for which we know the convergence/divergence, ${b}_{n}$:

If ${b}_{n} > {a}_{n}$ and ${b}_{n}$ converges, then ${a}_{n}$ also converges.
If ${b}_{n} < {a}_{n}$ and ${b}_{n}$ diverges, then ${a}_{n}$ also diverges.

We can compare this function to ${b}_{n} = \frac{1}{k} ^ 2$. We can do this because we know it converges (because of the p-test).

So, since $\frac{1}{k} ^ 2 > \frac{1}{9 {k}^{2}}$, and $\frac{1}{k} ^ 2$ converges, we can say that the series converges

But, wait, we only proved that this series converges when the numerator = 1. What about all the other values $\cos \left(\frac{1}{k}\right)$ could take? Well, remember that 1 is the maximum value that the numerator could take. So, since we have proven that this converges, we have indirectly proven that this series has converged for any value in the numerator.

Hope that helped :)