Integral Test for Convergence

Forum for the GRE subject test in mathematics.
huayualice
Posts: 15
Joined: Thu Sep 11, 2014 7:19 am

Integral Test for Convergence

Postby huayualice » Tue Oct 21, 2014 8:37 am

\sum_1^\infty \ln(1+1/n) is divergent.

I understand what the solution is saying (multiplying and eliminating everything except end terms....), but I couldn't figure out what I did wrong using the integral test: f(x) = ln (1+1/x) is decreasing, and positive, so \int_1^\infty \ln(1+1/x)dx = \int_1^\infty \ln(x+1)- lnx dx. Then letting y = x + 1 yields \int_2^\infty lnydy- \int_1^\infty lnx dx. It is integrating the same function from different starting points, won't the difference be a finite area?

Can anyone explain? Thanks!

Austin
Posts: 26
Joined: Fri Jun 01, 2012 9:32 am

Re: Integral Test for Convergence

Postby Austin » Tue Oct 21, 2014 10:05 am

Making use of the same logarithm property that you used, we can rewrite

\sum\limits_{n=1}^\infty \text{ln}\left(1+\frac{1}{n}\right) = \text{ln}\left( \prod\limits_{n=1}^\infty \left( 1+\frac{1}{n} \right) \right) > \text{ln}\left( \sum\limits_{n=1}^\infty \frac{1}{n} \right).

But this last sum is a p-series with p=1, and thus does not converge. The problem with the integral test in this case is that neither of the two integrals you finish with converges, and we can't very well say \infty-\infty will be some finite value. It's tempting to write

\int\limits_2^\infty \text{ln}(x)dx - \int\limits_1^\infty \text{ln}(x)dx = -\int\limits_1^2\text{ln}(x)dx,

but the familiar integral rewriting rules are only valid when all involved integrals converge.

I hope this helps!

huayualice
Posts: 15
Joined: Thu Sep 11, 2014 7:19 am

Re: Integral Test for Convergence

Postby huayualice » Wed Oct 22, 2014 2:12 am

Thanks! That helps.




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