Hi guys I've been doing some problems on series and I ran into this one,

which I have no idea how to go about doing.

Any help would be appreciated.

Thanks.

Printable View

- Nov 11th 2008, 11:29 AMHweengeeNeed help with convergence of this series
Hi guys I've been doing some problems on series and I ran into this one,

which I have no idea how to go about doing.

Any help would be appreciated.

Thanks. - Nov 11th 2008, 12:31 PMLaurent
Think about what the integral looks like when is large: then you integrate on a very small interval the function , which is almost equal to since is small... and you can integrate explicitly. It gives , which is the general term of a convergent series.

This justifies why the majoration will allow you to conclude. - Nov 11th 2008, 12:47 PMHweengee
So when n is large the interval [0, becomes small. And this implies that the area bounded by the curve becomes smaller and smaller, and so the series is likely to converge? Hence the comparison with a convergent series?

- Nov 11th 2008, 12:51 PMMathstud28
You could also do it this way

It should be clear that since the interval of integration is positive and so is the integrand that

We can also see that it is continuous (Ill try to prove this if you need be). And finally . So we may apply the integral test

Therefore the series shares the convergence/divergence of

Now switching the order of integration gives

All of which converge (for the justification of the first just consider that . Therefore the series converges. - Nov 11th 2008, 12:51 PMLaurent
Not only is the integral smaller and smaller, but it is like . The idea is to neglect the denominator since it becomes close to 1, so that we obtain a function that we know how to integrate. Actually I almost wrote the full proof. Just use the majoration I gave and integrate it.

- Nov 11th 2008, 12:52 PMHweengee
Ok thanks alot guys.