Does the series 4/k(lnk)^2 converge or diverge?

[sadcollegestudent]

Homework Statement

$\sum_{k=0}^\infty \frac 4 k(\ln k)^2$

Homework Equations

The Attempt at a Solution

I tried to solve it using the integral test but since it's not continuous it doesn't work.

 

[Greg Bernhardt]

I tried to solve it using the integral test but since it's not continuous it doesn't work.

Please show what you've tried

 

[fresh_42]

Homework Statement

$\sum_{k=0}^\infty \frac 4 k(\ln k)^2$

Homework Equations

The Attempt at a Solution

I tried to solve it using the integral test but since it's not continuous it doesn't work.

I've corrected your formula such that it is LaTeX compatible. Is that what you meant? And starting the summation with $0$?

 

[willem2]

∑∞k=04k(lnk)2∑k=0∞4k(ln⁡k)2\sum_{k=0}^\infty \frac 4 k(\ln k)^2

Are you sure \sum_{k=1}^\infty \frac 4 {k(\ln k)^2} wasn't meant? If you multiply with (ln k)^2 it's a rather simple summation if you know how to sum 1/k

 

[Ray Vickson]

Homework Statement

$\sum_{k=0}^\infty \frac 4 k(\ln k)^2$

Homework Equations

The Attempt at a Solution

I tried to solve it using the integral test but since it's not continuous it doesn't work.

Do you mean
$$\sum_k \frac{4}{k} (\ln k)^2 \hspace{3ex}(1)$$ or $$\sum_k \frac{4}{k (\ln k)^2} \hspace{3ex}(2) ?$$

Edit: I see that willem2 has beat me to it.

 

[Mark44]

Are you sure \sum_{k=1}^\infty \frac 4 {k(\ln k)^2} wasn't meant? If you multiply with (ln k)^2 it's a rather simple summation if you know how to sum 1/k

Based on what the OP wrote and revised a couple of times, this looks like what he/she intended.

Let's let @sadcollegestudent come back and let us know what the actual summation is...

 

[Ray Vickson]

Homework Statement

$\sum_{k=0}^\infty \frac 4 k(\ln k)^2$

Homework Equations

The Attempt at a Solution

I tried to solve it using the integral test but since it's not continuous it doesn't work.

(1) The integral test works perfectly. Perhaps if you showed us what you did we would be in a better position to help you.
(2) You need to start the summation at $k=2$, since the terms are undefined when $k = 0$ or $k = 1$ (assuming you have $\ln^2(k)$ in the denominator).