Divergence/convergence infinite series

[afkguy]

Homework Statement

show \sum 1/(ln k)^n diverges,for any n. the indexing is k = 2,3,...

Homework Equations

The Attempt at a Solution

Because k > ln k, k^n > ln k^n, and 1/k^n < 1/ln k^n

so this is just a p-series, which diverges for p =< 1.

So now I need to show it diverges for n > 1, but I'm having trouble finding a series to compare it with.

 

[hunt_mat]

What about using another method like the ratio test? Or the nth root test?

 

[afkguy]

I don't think either of those will work because the ratio test just gives the limit of a_k+1 over a_k = 1 and the kth root test doesn't seem to help because the limit also tends to 1.

Just to be clear, the indexing is from k = 2,3,... and this series is divergent for any n.

 

[afkguy]

ok so this problem became a lot easier once i found out k > (ln k)^n, for any k larger than
some K. If I use this fact, then the problem becomes easy as I can just compare 1/(ln k)^n to 1/k and remove the first 2 through K terms.

But, I have no idea how to show that k > (ln k)^n for big enough k?

 

[hunt_mat]

I use calculus. Examine the function:

<br /> f(x)=\frac{(\log x)^{n}}{x}<br />

Differentiate this function and find out that it is either increasing or decreasing and use that (with an initial value) to get the inequality for a variable x and then it much also be true for natural numbers.

 

[afkguy]

Ok, I think I see where to go.

so if f(x) = (lnx)^n/x then f'(x) = [nln(x)^(n-1) - 2ln(x)^n] /x^3

the denominator is positive on our domain [2,infinity), the numerator is negative when x > e^n/2

so f(x) is decreasing when x > e^n/2, call this K

plug in f(K) = some number, call it R

so f(x) is positive and decreasing, thus

R > f(x), when x > K.

so R > (lnx)^n/x, or 1/x < R/(lnx)^n, for x > K

if we consider the series and remove the first 2,...,K terms,

we have the series of 1/x < the series of R/(lnx)^n.

1/x diverges and so must R/(lnx)^n

R is just a finite # so if the series of R/(lnx)^n diverges, then the series of 1/(lnx)^n must as well

Did I follow you correctly?

 

[hunt_mat]

Yes.