# Problems proving a summation

• May 5th 2012, 01:45 PM
klw289
Problems proving a summation
I have been trying for a week to prove a result from a textbook and have gotten nowhere. Please help.

I have the facts $\displaystyle q^{-\delta}$$\displaystyle logq$$\displaystyle$\rightarrow$$\displaystyle \infty$$ for each$\displaystyle$\delta$>0$ and I'm trying to prove for each $\displaystyle \epsilon>0$
$\displaystyle \sum_{q=1}^{\infty}((logq)/(q^{1+\epsilon}))<\infty$
• May 5th 2012, 03:04 PM
Plato
Re: Problems proving a summation
Quote:

Originally Posted by klw289
I have been trying for a week to prove a result from a textbook and have gotten nowhere. Please help.

I have the facts $\displaystyle q^{-\delta}$$\displaystyle logq$$\displaystyle$\rightarrow$$\displaystyle \infty$$ for each$\displaystyle$\delta$>0$ and I'm trying to prove for each $\displaystyle \epsilon>0$
$\displaystyle \sum_{q=1}^{\infty}((logq)/(q^{1+\epsilon}))<\infty$

This is series of tricks. I am going to use easier notation.
Suppose that $\displaystyle p>1$ and let $\displaystyle r=\frac{p-1}{2}$.

Set some things up first. $\displaystyle \log(n)=r^{-1}\log(n^r)\le r^{-1}n^r$

So $\displaystyle \dfrac{\log(n)}{n^p}\le\dfrac{n^r}{r\cdot n^p}=\dfrac{1}{r\cdot n^{p-r}}$

If we note that $\displaystyle p-r=\frac{p+1}{2}>1$.

So you have a p-series which converges.