# series

• Feb 16th 2011, 12:21 AM
problem
series
Test the convergence of $\displaystyle \sum_{n=0}^\infty \frac{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k-1}}{\ln (k)}$.

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

Can anyone help me?
• Feb 16th 2011, 12:27 AM
chisigma
The first step in order to extablish if a series $\displaystyle \displaystyle \sum_{n=0}^{\infty} a_{n}$ converges or not is to verify the $\displaystyle \displaystyle \lim_{n \rightarrow \infty} a_{n}$. How can we say in this case?...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$
• Feb 16th 2011, 05:15 AM
CaptainBlack
Quote:

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

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:

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