Does the Series 1/(n * Log(n)) Converge Using the Cauchy Condensation Test?

[tarheelborn]

Homework Statement

I need to determine, using the Cauchy Condensation Test, whether or not
the series 1/(n * Log(n)) converges.

Homework Equations

The Attempt at a Solution

I believe that this series converges iff 2^n(1/(2^n*Log(2^n)) converges (Cauchy Condensation Test). I believe the series actually diverges, but I am not sure how to work through it. Thanks for your help.

 

[Office_Shredder]

Try simplifying those new terms that you got.

 

[gomunkul51]

Basically not hard at all, if you know that 1/n diverges.
use log rules. and eliminate doubles in nominator and denominator.

Also can be checked that it diverges by the Integral Test.

Good luck.

 

[tarheelborn]

Thank you so much!

 

[gomunkul51]

I have another one for you tarheelborn ! :)

"determine, using the Cauchy Condensation Test, whether or not
the series 1/[Log(n)^Log(n)] converges."

If you like you can try any other test !

*In Israel we never use Log(n) when we intend to say Ln(n) (as in Log in the natural base e), and in American literature they always get me confused with Log/Ln.
just a curiosity :)

 

[tarheelborn]

My professor used Log although he said that here is means Ln; I don't know quite why. I was accustomed to using Ln, too, in previous classes.

Meanwhile, to your problem... So 1/[Log(n)^Log(n)] converges iff 2^n*(1/[Log(2^n)^Log(2^n)] converges.

2^n*(1/[Log(2^n)^Log(2^n)] = 2^n/[Log(2^n)*Log(2^n)]
= 2^n/[n log(2) * n log(2)]
= 2 log(2) * Sum [2^n/n^2]
= 2 log(2) * Sum [1/n^2] which converges so
1/[Log(n)^Log(n)] converges.

Thanks for the challenge!

 

[gomunkul51]

Very good try !

but there is a little problem :)

"2^n*(1/[Log(2^n)^Log(2^n)] = 2^n/[Log(2^n)*Log(2^n)]"

you changed "Log(2^n)^Log(2^n)" to "Log(2^n)*Log(2^n)" so you basically did
ln(x)^a = a*ln(x), which isn't true..
ln(x^a) = a*ln(x) : this one is OK :)
have another try!

 

[tarheelborn]

Ok, here is my next attempt:

2^n*(1/[Log(2^n)^Log(2^n)] = 2^n/[n*Log(2)^(n*Log(2))]. Now 2^n goes to 1.
Log(2) is constant, so we are basically left with 1/n^n which is a quickly increasing version of 1/n which does not converge.

I am not sure if I can make those assumptions, but I don't know how else to deal with the Log. Again, thanks for the challenge; it is fun.

 

[System]

You have to use the cauchy's test?
If not: For an easier life, use the integral test.

 

[gomunkul51]

Another good try :)

Basically right, but 2^n does not go to 1, but you right that it grows much slower then n^n, so n^n is the dominate expression.

When you simplify the expression you get:

2^n / [n^(n*ln(2))*ln(2)^(n*ln(2))]

Which you can test by Cauchy's n'th Root Test to be convergent:

2 / [n^(ln(2))*ln(2)^(ln(2))] < 1

Another way altogether is to say that after some n: (e^2) is smaller then (ln(n)) :

1/[ln(n)^ln(n)] < 1/[(e^2)^ln(n)] = 1/[(e^(2*ln(n))] = 1/[(e^(ln(n^2))] = 1/[n^2] !
1/[n^2] is convergent. so the original expression is also convergent thorough direct comparison test :)

*Know your logarithmic rules ! :)

Nevermind, Good Luck to you !

 

[tarheelborn]

I do need to know the logarithmic rules more thoroughly! Sometimes I wonder if I will ever know all the rules and be able to remember them as needed. Practice, practice!

Thank you for this challenge.