1. ## Convergence Proof

if Sn is the summation of partial sums from 1 to infinity of (log n)^-logn, prove whether Sn converges or diverges.

It gives a hint of using the 2^m test which states that if Am is a nonincreasing function with limit 0 then the Summation of An from 1 to infinityconverges if and only if the series of the summation of partial sums 2^m*A(2^m) from m=0 to m=infinity

2. Originally Posted by Snooks02
if Sn is the summation of partial sums from 1 to infinity of (log n)^-logn, prove whether Sn converges or diverges.

It gives a hint of using the 2^m test which states that if Am is a nonincreasing function with limit 0 then the Summation of An from 1 to infinityconverges if and only if the series of the summation of partial sums 2^m*A(2^m) from m=0 to m=infinity
Yeah, just use Cauchy's Condensation test. Since $\frac{1}{\ln(n)^{\ln(n)}}$ is positive and nonincreasing we may apply Cauchy's Condenstaion test which states that $\sum_{a_n}$ converges iff $\sum{2^na_{2^n}}$ does. And in this case we have that $\sum{2^na_{2^n}}=\sum\frac{2^n}{(n\ln(2))^{n\ln(2) }}$

Now you can use whatever tes you want, Ratio, Root, comparison, whatever to determine that that converges.