1. ln convergence question

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

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 $\displaystyle u = x^{-\frac{1}{2}}$ and $\displaystyle dv = \frac{\ln{x}}{x}\,dx$

5. Re: ln convergence question

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

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

6. Re: ln convergence question

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

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

7. Re: ln convergence question

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

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

... and its derivative is...

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

... so that is...

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

Kind regards

$\chi$ $\sigma$