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Proving convergence of 1/log(n)

by hivesaeed4
Tags: 1 or logn, convergence, proving
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hivesaeed4
#1
Apr2-12, 11:14 PM
P: 217
Has anybody got any idea as to how to prove that Ʃ 1/(n(log(n))^p) converges? (where p>1)
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M Quack
#2
Apr3-12, 02:44 AM
P: 669
0 < log(n) < n for n>3
DonAntonio
#3
Apr3-12, 06:35 AM
P: 606
Quote Quote by hivesaeed4 View Post
Has anybody got any idea as to how to prove that Ʃ 1/(n(log(n))^p) converges? (where p>1)

First way, the integral test: [itex]\int_2^\inf \frac{1}{x\log^px} dx=\frac{\log^{1-p}(x)}{1-p}|_2^\inf \rightarrow \frac{log^{1-p}(2)}{p-1}[/itex] .

Second way, Cauchy's Condensation test: taking [itex]n=2^k[/itex] , the series's general term is [itex]\frac{1}{2^kk^p\log^p(2)}[/itex] , so multiplying this by [itex]2^k[/itex] we get [itex]\frac{1}{k^p\log^p(2)}[/itex] , which is a multiple of the series of [itex]\frac{1}{k^p}[/itex] , which we know converges for [itex]p>1[/itex] .

DonAntonio

hivesaeed4
#4
Apr3-12, 09:32 AM
P: 217
Proving convergence of 1/log(n)

Quote Quote by DonAntonio View Post
First way, the integral test: [itex]\int_2^\inf \frac{1}{x\log^px} dx=\frac{\log^{1-p}(x)}{1-p}|_2^\inf \rightarrow \frac{log^{1-p}(2)}{p-1}[/itex] .

Second way, Cauchy's Condensation test: taking [itex]n=2^k[/itex] , the series's general term is [itex]\frac{1}{2^kk^p\log^p(2)}[/itex] , so multiplying this by [itex]2^k[/itex] we get [itex]\frac{1}{k^p\log^p(2)}[/itex] , which is a multiple of the series of [itex]\frac{1}{k^p}[/itex] , which we know converges for [itex]p>1[/itex] .

DonAntonio
Thing is I've got to prove determine it's convergence from 1 to infinity.
hivesaeed4
#5
Apr3-12, 09:33 AM
P: 217
Just to clarify, the previous message was supposed to be:
Thing is I've got to determine it's convergence from 1 to infinity.
Office_Shredder
#6
Apr3-12, 09:37 AM
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If the sum starts at 1, you have a problem because log(1)=0
M Quack
#7
Apr3-12, 11:02 AM
P: 669
The sum from 1 to any fixed number will always converge (unless you divide by zero).
Therefore, if the sum from any fixed number to infinity converges, then the sum from 1 to infinity converges.

You can choose any fixed number you like. I chose 3 because 0<log(n) < n for all n>=3.

You can choose 11 if your log is base 10 rather than natural, or because, you know, usual amps only go up to 10, but yours goes all the way up to 11.

The sum from 3 to infinity of your series is smaller than the sum from 3 to infinity over 1/n^(p+1), which is smaller than 1/n^2 which we know converges.


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