Is the Series \(\sum_{n=1}^\infty \frac{\cos{\frac{n}{2}}}{n^2+4n}\) Convergent?

[Townsend]

\sum_{n=1}^\infty\frac{\cos{\frac{n}{2}}}{n^2+4n}


Sorry I am just trying out this latex stuff as it's very new to me.

Anyways, I want to test a series for convergence. The series is (if the latex does not work right) suppose to be

The sum from n=1 to infinity [cos(n/2) / (n^2+4n)]

Since this series has postive and negative terms but not alternating terms I have a limited number of test to try.

I used the Ratio test.

So I take lim as n goes to infinity of [cos((n+1)/2) / ((n+1)^2+4(n+1))] over
[cos(n/2) / (n^2+4n)]

and of course that is in absolute value bars.

Now as n goes to infinity (n^2+4n)/((n+1)^2+4(n+1)) goes to one and I am left with the limit as n goes to inifinity of

Absolute value[ cos((n+1)/2) / cos(n/2) ]

Now I am sure this goes to one since my calculator can take this limit, but how could someone actually take this limit is another question. Can anyone help?

Thanks

P.S. I will work on this latex so maybe next time things will look better

 

[arildno]

Hint:
Sum(i from 1 to inf)1/(i^(2)) is convergent

 

[M98Ranger]

Your approach using the Ratio Test is correct. As you have shown, the limit of the absolute value of the ratio of consecutive terms approaches 1, indicating that the series is inconclusive. In this case, the Ratio Test is not able to determine the convergence or divergence of the series.

Other tests that could be used to test for convergence include the Comparison Test, the Limit Comparison Test, and the Alternating Series Test. However, since the series does not have alternating terms and may not be easily comparable to other known series, these tests may not be effective in this case.

In situations like this, it may be helpful to use a graphing calculator or software to plot the terms of the series and observe the behavior. From the graph, it appears that the terms oscillate around 0 and decrease in magnitude as n increases, suggesting that the series may converge. However, this is not a conclusive proof and further analysis would be needed to determine the convergence or divergence of the series.