question on divergence of a series?

Question

hi. suppose 0
how to show that sum from n=2 to infinity of a_n diverges, where a_n is defined as

a_n =
(2*c*log(n)-1) divided by
[(n^c)*(log(n))^(3/2)]

thanks.

Answer 1

What a mess!
0 < c < 1.
What I'm going to show is that for n large enough the quotient is > 1/n whose series is known to diverge. So by comparison the given series diverges.
The numerator: 2clog(n) - 1; since c is a constant > 0 and
log(n) goes to infinity, there is an N such that for all n > N
2clog(n) > 2 and the numerator is > 1.
The denominator: (n^c)*(log(n))^3/2. Want to show this is less than n eventually by dividing it by n and use L'Hopital's
rule to show the quotient goes to zero. I did it using log base e, though that doesn't make any difference except for constants. You have to do it twice, but it works, the quotient goes to 0.
Okay,for your original quotient, the num >1 eventually and denominator < n eventually so eventually the terms are
>1/n.
If you need more help with this: [e-mail]

Answer 2

the series is of the same class as 1/n^c*(log n) ^1/2>
1/nln(n) which is divergent
To show that n ln n is divergent
take y =ln(ln(x) and apply Lagrange to the interval n,n+1
ln ln(n+1) -ln(ln n) = 1/(n+a)ln(n+a) 0 so ln(ln(n+1)+ln(ln(n) <1/n ln n
sum both sides from n=2
ln(ln(n+1) -ln(ln(2) < sum 1/nln n
as the left side ==> infinity so does the right side so a_n = 1/n ln n is divergent