If $\pi_k(n)$ is the cardinality of numbers with k factors (repetitions included) less than or equal n, the generalized Prime Number Theorem is:
$\pi_k(n)\sim \frac{n}{\ln n} \frac{(\ln \ln n)^{k-1}}{(k-1)!}.$
I noticed it appears true that
$\lim_{n \to\infty}\ \sum_{k=1}^{n}\frac{2^n}{\ln 2^n} \frac{(\ln\ln 2^n)^{k-1}}{(k-1)!} = 2^n ,$
which makes sense to me. In my attempts to prove this I could only get a few steps along. How can this be done?
Edit: Looking through Ramanjuan's Collected Papers in the 32d paper I notice he has the following:
$[x] = \{\pi_1(x) + \pi_2(x)+\pi_3(x)...\}.......(1)$
and
$x = \frac{x}{\ln x}\{1 + \ln\ln x + \frac{(\ln\ln x)^2}{2!}...\}....(2)$
He says that (1) and (2) are "obvious." The second was not obvious to me, but can be found by letting y = $\ln\ln x$ and using the Taylor series for e. I'm including this for completeness because (1) above is the idea behind the original question.