# 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 $\sum{a_n}$ be an infinite series with $a_{n+1}\leqslant{a_n}$ and $a_n>0$. Then $\sum{a_n}$ converges iff $\sum{2^na_{2^n}}$ converges. So since $\forall{k}\in(2,\infty)~\frac{\log(k)}{k^p}>1$ and monotonically decreasing we may apply the above test. So

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

\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 $\limsup\sqrt[n]{|a_n|}<1$

So solving this for p gives $p>1$

3. Originally Posted by thahachaina
Hey guys,

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

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

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

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

CB