# sums with lnx

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• Apr 21st 2008, 12:53 PM
akhayoon
sums 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 PM
Jhevon
Quote:

Originally Posted by akhayoon
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 $\displaystyle \sum\frac{lnk}{k^{2}}$

try the integral test
• Apr 21st 2008, 01:01 PM
Moo
Hello,

I think ratio test can work too...
• Apr 21st 2008, 01:02 PM
Mathstud28
Quote:

Originally Posted by Jhevon
try the integral test

(Thinking)...the first one would yield an Ei(x) function...........................thats gross.........or am I integrating wrong? (Rofl)....
• Apr 21st 2008, 01:09 PM
Jhevon
Quote:

Originally Posted by Moo
Hello,

I think ratio test can work too...

the ratio test is inconclusive

Quote:

Originally Posted by Mathstud28
(Thinking)...the first one would yield an Ei(x) function...........................thats gross.........or am I integrating wrong? (Rofl)....

you can use the integral test for the second and then use the comparison test for the first in comparing the second with the first
• Apr 21st 2008, 01:12 PM
Mathstud28
Quote:

Originally Posted by Jhevon
the ratio test is inconclusive

you can use the integral test for the second and then use the comparison test for the first in comparing the second with the first

(Rofl)...ahh...I thoguth you just meant to do the integral test....(Giggle)
• Apr 21st 2008, 01:13 PM
Jhevon
Quote:

Originally Posted by Mathstud28
(Rofl)...ahh...I thoguth you just meant to do the integral test....(Giggle)

yes, my apologies. I am a man of few words (except in MHF chat :D) and so it is easy to misunderstand me....
• Apr 21st 2008, 01:52 PM
Plato
$\displaystyle \ln (n) \leqslant 2\sqrt n \Rightarrow \quad \frac{{\ln (n)}} {{n^2 }} \leqslant \frac{2} {{n^{3/2} }}$
Simple comparison with p-series.
• Apr 21st 2008, 02:00 PM
akhayoon
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 PM
Mathstud28
Quote:

Originally Posted by akhayoon
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)

He just picked a term that was greater than ln(x)...which is $\displaystyle 2\sqrt{x}$...and he got that from the fact that $\displaystyle \frac{\sqrt{x}}{x^2}=x^{-2+\frac{1}{2}}=x^{\frac{-3}{2}}$
• Apr 21st 2008, 02:03 PM
akhayoon
so that was completely arbitrary and u could've essentially picked anything greater than lnx??
• Apr 21st 2008, 02:03 PM
Jhevon
Quote:

Originally Posted by Mathstud28
He just picked a term that was greater than ln(x)...which is $\displaystyle 2\sqrt{x}$...and he got that from the fact that $\displaystyle \frac{\sqrt{x}}{x^2}=x^{-2+\frac{1}{2}}=x^{\frac{-3}{2}}$

i think the posters point is, how did he know what to pick?
• Apr 21st 2008, 04:51 PM
akhayoon
well how would you even integrate $\displaystyle \int \frac{lnx}{e^{x}}$

by parts?
• Apr 21st 2008, 05:00 PM
Mathstud28
Quote:

Originally Posted by akhayoon
well how would you even integrate $\displaystyle \int \frac{lnx}{e^{x}}$

by parts?

You can't do it..well not by normal means...by power series you coud...but to be blun $\displaystyle \int\frac{\ln(x)}{e^{x}}dx=\frac{-\ln(x)}{e^{x}}-Ei_1(x)$ where Ei(x) is the exponential integral is $\displaystyle E_1(x)=\int_x^{\infty}\frac{e^{-t}}{t}dt$
• Apr 21st 2008, 05:06 PM
akhayoon
$\displaystyle \sum\frac{lnk}{e^{k}}$

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..
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