Series convergent or divergent?

[renegade05]

I am working with the following series:

\(\displaystyle \sum_{n=1}^{\infty}\frac{1}{1+\ln{n}}\)

The question states: For the following series, determine (with analytical reasons) whether the series converges or diverges. State the test you use. If convergent, find the sum.

I am thinking comparison test?

\(\displaystyle 1+\ln(n)<n \) when n>1

So \(\displaystyle \frac{1}{1+\ln{n}} > \frac{1}{n}\)

And since \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{n}\) diverges (harmonic series) than \(\displaystyle \sum_{n=1}^{\infty}\frac{1}{1+\ln{n}}\) also diverges.

Is this reasoning sound? Or am I (confused)?

P.S. I like the new forum design.

 

[galactus]

Yes, your reasoning seems sound. The series diverges.

But, the clue was when it asked to find the sum if it converged.

If this converged, it would be difficult to find a closed form.

Thus, it diverges.

But seriously, take \(\displaystyle \displaystyle\sum_{n=2}^{\infty}\frac{1}{nln(n)}\). This diverges because it behaves like the harmonic series.

If its terms were larger than the harmonic series, we would expect it to diverge. But, they are smaller.
So, using the integral test on this one we find it diverges.

We could use the integral test on yours as well. It is tough to integrate by hand. But running it through a calculator, we get \(\displaystyle \displaystyle\int_{1}^{\infty}\frac{1}{1+ln(x)}dx=\infty\)
Thus, divergent.

 

[renegade05]

We could use the integral test on yours as well. It is tough to integrate by hand. But running it through a calculator, we get

Unfortunately, this question appeared in the no-calculator portion of my exam. I got it wrong and my teacher doesn't post solutions (stupid) so I was just wondering the correct answer.

 

[galactus]

Yes, it is divergent. That is correct.