series

**problem** · February 16th 2011, 12:21 AM

Test the convergence of $\sum_{n=0}^\infty \frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}}{\ln (k)}$ .

I try to use ratio test but I am stuck in summing up the numerator part.

Can anyone help me?

**chisigma** · February 16th 2011, 12:27 AM

The first step in order to extablish if a series $\displaystyle \sum_{n=0}^{\infty} a_{n}$ converges or not is to verify the $\displaystyle \lim_{n \rightarrow \infty} a_{n}$ . How can we say in this case?...

Kind regards

$\chi$ $\sigma$

**CaptainBlack** · February 16th 2011, 05:15 AM

Originally Posted by problem

Test the convergence of $\sum_{n=0}^\infty \frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}}{\ln (k)}$ .

I try to use ratio test but I am stuck in summing up the numerator part.

Can anyone help me?

You might consider making the summand depend on the summation variable:

$\displaystyle \sum_{k=0}^\infty \frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}}{\ln (k)}$ .

Now re-read chisigma's post.

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