sums with lnx

**akhayoon** · April 21st 2008, 12:53 PM

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

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

**Jhevon** · April 21st 2008, 01:00 PM

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

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

try the integral test

**Moo** · April 21st 2008, 01:01 PM

Hello,

I think ratio test can work too...

**Mathstud28** · April 21st 2008, 01:02 PM

Originally Posted by Jhevon

try the integral test

...the first one would yield an Ei(x) function...........................thats gross.........or am I integrating wrong?

....

**Jhevon** · April 21st 2008, 01:09 PM

Originally Posted by Moo

Hello,

I think ratio test can work too...

the ratio test is inconclusive

Originally Posted by Mathstud28

...the first one would yield an Ei(x) function...........................thats gross.........or am I integrating wrong?

....

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

**Mathstud28** · April 21st 2008, 01:12 PM

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

...ahh...I thoguth you just meant to do the integral test....

**Jhevon** · April 21st 2008, 01:13 PM

Originally Posted by Mathstud28

...ahh...I thoguth you just meant to do the integral test....

yes, my apologies. I am a man of few words (except in MHF chat

) and so it is easy to misunderstand me....

**Plato** · April 21st 2008, 01:52 PM

$\ln (n) \leqslant 2\sqrt n \Rightarrow \quad \frac{{\ln (n)}}<br /> {{n^2 }} \leqslant \frac{2}<br /> {{n^{3/2} }}$
Simple comparison with p-series.

**akhayoon** · April 21st 2008, 02:00 PM

I don't understand how that comparison works...I mean where did the $2\sqrt{n}$ come from and how did it become to the exponent 3/2?? so confused

**Mathstud28** · April 21st 2008, 02:02 PM

Originally Posted by akhayoon

I don't understand how that comparison works...I mean where did the $2\sqrt{n}$ come from and how did it become to the exponent 3/2?? so confused

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

**akhayoon** · April 21st 2008, 02:03 PM

so that was completely arbitrary and u could've essentially picked anything greater than lnx??

**Jhevon** · April 21st 2008, 02:03 PM

Originally Posted by Mathstud28

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

i think the posters point is, how did he know what to pick?

**akhayoon** · April 21st 2008, 04:51 PM

well how would you even integrate $\int \frac{lnx}{e^{x}}$

by parts?

**Mathstud28** · April 21st 2008, 05:00 PM

Originally Posted by akhayoon

well how would you even integrate $\int \frac{lnx}{e^{x}}$

by parts?

You can't do it..well not by normal means...by power series you coud...but to be blun $\int\frac{\ln(x)}{e^{x}}dx=\frac{-\ln(x)}{e^{x}}-Ei_1(x)$ where Ei(x) is the exponential integral is $E_1(x)=\int_x^{\infty}\frac{e^{-t}}{t}dt$

**akhayoon** · April 21st 2008, 05:06 PM

$\sum\frac{lnk}{e^{k}}$

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

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

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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