Using the integral test, how do you show whether #sum (1/n^2)cos(1/n) # diverges or converges from n=1 to infinity?

Question

Emilia Aguilar · Accepted Answer

The #cos(1/n)# factor is a bit of a distraction.

#sum_(n=1)^oo 1/(n^2)# is absolutely convergent (as can be easily shown by a variety of means) and #abs(cos(1/n)) <= 1#.

Note that #int_1^oo 1/(x^2) dx = -1/x |_1^oo = 0 - (-1) = 1# So by the integral test #sum_(n=1)^oo (1/(n^2))# is convergent Now #1/(n^2) > 0# for all #n > 0#, so #sum_(n=1)^oo abs(1/(n^2))# is convergent. #abs(cos(1/x)) <= 1 AA x > 0# Therefore: #sum_(n=1)^oo abs(1/(n^2) cos(1/n)) <= sum_(n=1)^oo abs(1/(n^2))# is a sum of non-negative terms, bounded above and therefore convergent. Hence: #abs(sum_(n=1)^oo (1/(n^2)) cos(1/n)) <= sum_(n=1)^oo abs(1/(n^2) cos(1/n))# is convergent.

Elijah Abram · Answer

If you insist on using the integral test, you need to find #int_1^oo 1/x^2 cos(1/x) dx = lim_(brarroo)int_1^b 1/x^2 cos(1/x) dx  #
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Use substitution with #u = 1/x# to get #du = -1/x^2 dx# so the integral becomes:
#lim_(brarroo) -int_1^(1/b)sinu du = lim_(brarroo) -sin(1/b)-(-sin(1)) = sin1# The integral converges, so the series also converges. (I think it is obvious that the function #1/x^2 cos (1/x)# is eventually decreasing.  Although your grader might want you to show it.)

Elijah Abram · Answer

undefined To determine whether the series $ \sum \frac{1}{n^2} \cos\left(\frac{1}{n}\right) $ converges or diverges, we can use the integral test.

1. We start by considering the function $ f(x) = \frac{1}{x^2} \cos\left(\frac{1}{x}\right) $.

2. We need to check if $ f(x) $ is continuous, positive, and decreasing for all $ x $ from 1 to infinity.

3. The function $ f(x) $ is continuous and positive for $ x \geq 1 $.

4. To check if $ f(x) $ is decreasing, we examine its derivative $ f'(x) $.

5. Calculating the derivative $ f'(x) $, we find that it is not straightforward due to the product rule and the chain rule. However, we can observe that the cosine function oscillates between -1 and 1 as $ \frac{1}{x} $ oscillates between 1 and 0, resulting in alternating signs for the terms of the series.

6. Since $ \frac{1}{x^2} $ decreases as $ x $ increases, and the oscillation of $ \cos\left(\frac{1}{x}\right) $ does not affect the decreasing behavior of $ f(x) $ significantly, we can conclude that $ f(x) $ is decreasing for $ x \geq 1 $.

7. Now, we proceed to evaluate the integral $ \int_{1}^{\infty} f(x) \, dx $.

8. This integral represents the area under the curve of $ f(x) $ from 1 to infinity.

9. Integrating $ f(x) $ analytically may not be feasible due to the complexity of the cosine function within the integral.

10. However, we can estimate the integral numerically using computational methods or software.

11. If the integral $ \int_{1}^{\infty} f(x) \, dx $ converges, then the series $ \sum \frac{1}{n^2} \cos\left(\frac{1}{n}\right) $ converges by the integral test. Otherwise, if the integral diverges, the series also diverges.

Therefore, to determine the convergence or divergence of the series $ \sum \frac{1}{n^2} \cos\left(\frac{1}{n}\right) $ from $ n = 1 $ to infinity, we need to evaluate the integral $ \int_{1}^{\infty} \frac{1}{x^2} \cos\left(\frac{1}{x}\right) \, dx $.