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The Cauchy condensation test says that \[\sum_{n=1}^{\infty} f(n)\] converges if and only if \[\sum_{n=0}^{\infty} 2^n f(2^n)\] converges.

The sum given to evaluate is \[\sum_{n=1}^{\infty} \frac{1}{n^p \ln n}\]
This sum converges if and only if the sum \[\sum_{n=0}^{\infty} \frac{2^n}{(2^n)^p \ln (2^n)}\] converges.
Algebraically, the sum is equal to \[\sum_{n=0}^{\infty} \frac{2^n}{n(2^{np}) \ln 2}\]

We apply the ratio test.

\begin{eqnarray*}
L &=& \lim_{n\to\infty} |\frac{2^{n+1}}{(n+1) 2^{np+p} \ln 2} \frac{n(2^{np}) \ln 2}{2^n}|\\
&=& \lim_{n\to\infty} |\frac{2^{n+1}}{2^n} \frac{n}{n+1} \frac{2^{np}}{2^{np+p}}|\\
&=& |2^{1-p}|
\end{eqnarray*}

To find values of $p$ for which the original sum converges, we set the inequality $2^{1-p} < 1$ which gives $p > 1$ proving that the sum does not converge for $0 \le p < 1$.
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