How to prove that the series is converge?

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

Answer 1

Converges by the Direct Comparison Test.

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

#sum_(n=1)^oocos(1/k)/(9k^2)#, IE, the series starts at one.
To use the Direct Comparison Test, we have to prove that #a_k=cos(1/k)/(9k^2)# is positive on #[1,oo)#.
First, note that on the interval #[1, oo), cos(1/k)# is positive. For values of #x=1, 1/k
Furthermore, we can say #cos(1/k)<=1#, as #lim_(k->oo)cos(1/k)=cos(0)=1#.

Then, we can define a new sequence

#b_k=1/(9k^2)>=a_k# for all #k.#

Well,

#sum_(k=1)^oo1/(9k^2)=1/9sum_(k=1)^oo1/k^2#
We know this converges by the #p-#series test, it is in the form #sum1/k^p# where #p=2>1#.

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

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Answer 2

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

Recognize that the range of cosine is [-1,1]. Check out the graph of #cos(1/x)#:

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:

#sum1/(9k^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 = 1/k^2#. We can do this because we know it converges (because of the p-test).
So, since #1/k^2 > 1/(9k^2)#, and #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(1/k)# 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 :)

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Answer 3

To prove that a series converges, you can use various convergence tests such as the comparison test, ratio test, root test, integral test, or alternating series test. Each test has its conditions and method of application. You need to determine which test is applicable to the given series based on its characteristics. Then, follow the steps of the chosen test to demonstrate convergence. If the conditions of the test are met and the series satisfies the convergence criterion, you can conclude that the series converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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