Question

Let $0<p<1$. Show that $0\le \frac{1}{k\ln k}\le \frac{1}{k^p}$for large values of k. Explain why we cannot use a p-series with$0<p<1$in a comparison test to verify the divergence of the series$\sum _{k=2}^{\infty }\frac{1}{k\ln k}$.

Short answer

The comparison test provides no information on the series$\sum _{k=2}^{\infty }\frac{1}{k\left(\ln k\right)}$ divergence.

Step-by-step solution

Step 1. Given information

An expression is given as$0\le \frac{1}{k\ln k}\le \frac{1}{k^p}$

Step 2. Verification

For larger values of k ,

$$\begin{gathered} k^p\le k \\ {k}^P\ln k\le k\ln k \\ \frac{1}{k\ln k}\le \frac{1}{k^p\ln k} \\ \frac{1}{k^p\ln k}<\frac{1}{k^p} \\ \frac{1}{k\ln k}<\frac{1}{k^p} \end{gathered}$$

For $0<p<1$, the p-series is divergent. If the series $\sum _{k=1}^x{b}_k$converges, then the series $\sum _{i=1}^{\infty }{a}_k$converges, according to the comparison test.

But the series $\sum _{i=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^p}$is divergent, so we cannot find the value of convergence.

The comparison test provides no information on the series' divergence.