Calculus 2 infinite Series

[GreenPrint]

Homework Statement

Determine if the following series converges or diverges.

Ʃ[k=1,inf] tan(k)/(k^2+1)

Homework Equations

The Attempt at a Solution

I have no idea how to solve this problem. Now that I think of it, I have never solved a single question about series were I'm asked about convergence or divergence of a series with tangent or cotangent as part of the series. Tangent and cotangent are not defined at multiples of pi/2 excluding multiples of pi, but the series is from k to infinity were k is the set of integers and so the numerator all by itself would never go to positive or negative infinity at any k. Yet I can't seem to come up with a solution to this problem.

Also just a quick question. If I'm given a particular series in which I know the function which it represents, if the function is undefined at some given points, like for example 1/(2-x) or something of the sort, could I automatically conclude that the series doesn't converge at positive 2 sense the function doesn't?

Also I have never seen a problem were the interval of convergence included two intervals like [-10,5)(5,22] or something of the sort just [-10,5) if that makes any sense at all. Is it possible to have series were there are two intervals of convergence instead of just one?

Thank you for any help

 

[GreenPrint]

=( ah man

 

[Mark44]

I posted a message about this series in the Homework Helpers' section. Maybe somebody there will have a good idea. The integral test seems like a way to go, but I haven't carried it all the way through.

 

[I like Serena]

I believe this series diverges.

A condition for a series to converge is that it needs to be possible for any epsilon > 0, that there is an N'th term such that all subsequent terms have an absolute value less than epsilon.

tan(k) can get arbitrarily large as k comes arbitrarily close to pi/2 + m pi for some m.
However large you want it to be, you can get it for some k.
So you can make always find a term larger than epsilon.

However, I'm afraid my argument isn't completely rigorous yet.