I'm having trouble trying to figure out what to do for convergence and divergence for a series that has lnx in it

for example

$\displaystyle \sum\frac{lnk}{e^{k}}$ and $\displaystyle \sum\frac{lnk}{k^{2}}$

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- Apr 21st 2008, 12:53 PMakhayoonsums with lnx
I'm having trouble trying to figure out what to do for convergence and divergence for a series that has lnx in it

for example

$\displaystyle \sum\frac{lnk}{e^{k}}$ and $\displaystyle \sum\frac{lnk}{k^{2}}$ - Apr 21st 2008, 01:00 PMJhevon
- Apr 21st 2008, 01:01 PMMoo
Hello,

I think ratio test can work too... - Apr 21st 2008, 01:02 PMMathstud28
- Apr 21st 2008, 01:09 PMJhevon
- Apr 21st 2008, 01:12 PMMathstud28
- Apr 21st 2008, 01:13 PMJhevon
- Apr 21st 2008, 01:52 PMPlato
$\displaystyle \ln (n) \leqslant 2\sqrt n \Rightarrow \quad \frac{{\ln (n)}}

{{n^2 }} \leqslant \frac{2}

{{n^{3/2} }}$

Simple comparison with p-series. - Apr 21st 2008, 02:00 PMakhayoon
I don't understand how that comparison works...I mean where did the $\displaystyle 2\sqrt{n}$ come from and how did it become to the exponent 3/2?? so confused (Doh)

- Apr 21st 2008, 02:02 PMMathstud28
- Apr 21st 2008, 02:03 PMakhayoon
so that was completely arbitrary and u could've essentially picked anything greater than lnx??

- Apr 21st 2008, 02:03 PMJhevon
- Apr 21st 2008, 04:51 PMakhayoon
well how would you even integrate $\displaystyle \int \frac{lnx}{e^{x}}$

by parts? - Apr 21st 2008, 05:00 PMMathstud28
- Apr 21st 2008, 05:06 PMakhayoon
$\displaystyle \sum\frac{lnk}{e^{k}}$

so the book is telling me to use the ratio test...

and you get$\displaystyle \frac{ln(k+1)}{elnk}=\frac{1}{e}$

so now I guess how would ln(k+1)/lnk go to 1 would be my basic question...

all I know is is that ln(n)/n goes to 0 as n goes to infinity but i haven't seem something like this before..