1. ## ln convergence question

$\displaystyle \sum_{1}^{\infty}\frac{ln k}{k\sqrt {k}}$

how to shpw if it converges or diverges

2. ## Re: ln convergence question

Try the integral test. The integral is convergent.

3. ## Re: ln convergence question

but i have trouble with the integral
lnk=t
1/k=dt

but in the denominator another sqrt(k)

4. ## Re: ln convergence question

You need to use integration by parts with $\displaystyle \displaystyle u = x^{-\frac{1}{2}}$ and $\displaystyle \displaystyle dv = \frac{\ln{x}}{x}\,dx$

5. ## Re: ln convergence question

Originally Posted by transgalactic
$\displaystyle \sum_{1}^{\infty}\frac{ln k}{k\sqrt {k}}$ converges or diverges
This can be done by direct comparison.

Can you show that $\displaystyle \ln (k) \leqslant 4\sqrt[4]{k}~?$

6. ## Re: ln convergence question

Originally Posted by transgalactic
$\displaystyle \sum_{1}^{\infty}\frac{ln k}{k\sqrt {k}}$

how to shpw if it converges or diverges
$\displaystyle \log(k)$ goes to infinity more slowly than any power of $\displaystyle k$ so for $\displaystyle k$ large enough there exists a positive constant $\displaystyle A$ such that $\displaystyle \log(k)/k^{1.5}<A/k^{1.4} ...$
CB

7. ## Re: ln convergence question

For $\displaystyle \text{Re}\ s>1$ the 'Riemann zeta function' is defined as...

$\displaystyle \zeta(s)= \sum_{k=1}^{\infty} \frac{1}{k^{s}}$ (1)

... and its derivative is...

$\displaystyle \zeta^{'}(s)= - \sum_{k=1}^{\infty} \frac{\ln k}{k^{s}}$ (2)

... so that is...

$\displaystyle \sum_{k=1}^{\infty} \frac{\ln k}{k^{\frac{3}{2}}}= - \zeta^{'}(\frac{3}{2})$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$