convergence of a series sum of ln k / (k ^ (3/2)

[Dom]

Hi Gals and Guys,
can anyone help me on a series problem I am getting stuck on. Does the series , summation of ln k/ (k ^ (3/2)) converge or diverge? I know that the divergent harmonic series 1/k = sqrt k/ (k ^ (3/2) is greater than ln k / ( (k ^ (3/2) because numerators, sqrt k is greater than ln k. I also know that ln k / ( k ^ ( 3/2)) is greater than the convergent series sqrt k / (k ^ 2) ( sqrt k / (k ^2) = 1 / ( k ^ ( 3/2)) and this is a convergen p-series with 3/2 which is greater than 1 ).

Thank you in advance for any assistance.

Dom M

 

[pka]

. Does the series , summation of ln k/ (k ^ (3/2)) converge or diverge?

$\displaystyle \ln (k) = 4\ln \left( {\sqrt[4]{k}} \right) \leqslant 4{k^{\frac{1}{4}}}$. Now use simple comparison with a ​p-series.

 

[lookagain]

Does the series, the summation from 1 to infinity of ln k/ (k ^ (3/2)), converge or diverge? / / / I know that the divergent harmonic series, the sum from 1 to infinity of 1/k, = sqrt k/ (k ^ (3/2) is greater than ln k / ( (k ^ (3/2) because numerators, sqrt k is greater than ln k. I also know that the series sum from 1 to infinity of ln k / ( k ^ ( 3/2)) is greater than the convergent series sum from 1 to infinity of sqrt k / (k ^ 2). / / / ( sqrt k / (k ^2) = 1 / ( k ^ ( 3/2)) and this is a convergent p-series with 3/2 which is greater than 1 .

$\displaystyle Edit: \ \ \$ $\displaystyle \ \ \ \dfrac{ln(k)}{k^{3/2}}, \ \dfrac{k}{k^{3/2}}, \ \dfrac{1}{k}, \ etc.$ are general terms of their respective series, not the series themselves.

 

[pka]

$\displaystyle \ \ \ \dfrac{k}{k^{3/2}}, \ \dfrac{1}{k}, \ etc.$ are terms of their respective series, not the series themselves.

The question is about the series $\displaystyle \sum\limits_{k = 1}^\infty {\dfrac{{\ln (k)}}{{\sqrt {{k^3}} }}}$.

 

[lookagain]

The question is about the series $\displaystyle \sum\limits_{k = 1}^\infty {\dfrac{{\ln (k)}}{{\sqrt {{k^3}} }}}$.

Yes, I know that. To be clearer, I amended (edited) my previous post.

 

[Dom]

Thank you

Thank you for the assistance, you made it look easy :wink: