# How to prove that the series is converge?

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#sum (cos (1/k))/(9k^2)#

##### 2 Answers

Converges by the Direct Comparison Test.

#### Explanation:

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

To use the Direct Comparison Test, we have to prove that

First, note that on the interval

Furthermore, we can say

Then, we can define a new sequence

Well,

We know this converges by the

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

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

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:

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

Consider an arbitrary series

If

If

We can compare this function to

So, since **series converges**

But, wait, we only proved that this series converges when the numerator = 1. What about all the other values **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 :)