# Thread: Analysis (Series Convergence)

1. ## Analysis (Series Convergence)

Show that the SERIES log k / (k^p), p>1 converges.
thanks.

2. Originally Posted by thahachaina
Show that the SERIES log k / (k^p), p>1 converges.
thanks.
Let $\displaystyle \sum{a_n}$ be an infinite series with $\displaystyle a_{n+1}\leqslant{a_n}$ and $\displaystyle a_n>0$. Then $\displaystyle \sum{a_n}$ converges iff $\displaystyle \sum{2^na_{2^n}}$ converges. So since $\displaystyle \forall{k}\in(2,\infty)~\frac{\log(k)}{k^p}>1$ and monotonically decreasing we may apply the above test. So

$\displaystyle \sum2^n\frac{\log(2^n)}{\left(2^n\right)^p}=\sum\f rac{n\log(2)}{2^{n(p-1)}}$. Now applying the Root test gives

\displaystyle \begin{aligned}\limsup\sqrt[n]{\left|\frac{n\log(2)}{2^{n(p-1)}}\right|}&=\limsup\frac{\sqrt[n]{n\log(2)}}{2^{p-1}}\\ &=\frac{1}{2^{p-1}}\end{aligned}

So now for a series to converge we must have that $\displaystyle \limsup\sqrt[n]{|a_n|}<1$

So solving this for p gives $\displaystyle p>1$

3. Originally Posted by thahachaina
Hey guys,

Does the SERIES, log k/ k^p, where p>1, converge or diverge???

Thanks
It converges because $\displaystyle \log(k)$ increases more slowly than $\displaystyle k^{(p-1)/2}$ and so eventually:

$\displaystyle \frac{\log(k)}{k^p}<\frac{1}{k^{\frac{p+1}{2}}}=\f rac{1}{k^q}$

and $\displaystyle q=(p+1)/2>1$.

CB