$$\sum_1^\infty \ln(1+1/n)$$ 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!
Integral Test for Convergence
Re: Integral Test for Convergence
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 pseries 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!
$$\sum\limits_{n=1}^\infty \text{ln}\left(1+\frac{1}{n}\right) = \text{ln}\left( \prod\limits_{n=1}^\infty \left( 1+\frac{1}{n} \right) \right) > \text{ln}\left( \sum\limits_{n=1}^\infty \frac{1}{n} \right).$$
But this last sum is a pseries 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!

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 Joined: Thu Sep 11, 2014 7:19 am
Re: Integral Test for Convergence
Thanks! That helps.