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Townsend
Apr28-04, 11:34 PM
\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
Apr29-04, 02:22 AM
Hint:
Sum(i from 1 to inf)1/(i^(2)) is convergent