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:

$$[x] = \{\pi_1(x) + \pi_2(x)+\pi_3(x)...\}.......(1)$$


$$x = \frac{x}{\ln x}\{1 + \ln\ln x + \frac{(\ln\ln x)^2}{2!}...\}....(2)$$

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.

  • $\begingroup$ the right hand side depends on $n$ :-) $\endgroup$
    – vesszabo
    Aug 9 '12 at 20:58

Consider, for some constant $\kappa$, $$ \begin{eqnarray} c_n(\kappa) &=& \frac{1}{n \kappa }\sum_{k=0}^{n-1} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!} = \frac{1}{n \kappa } \left( \sum_{k=0}^{\infty} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!} - \sum_{k=n}^{\infty} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!} \right) \\ \ & =& 1 - \frac{1}{n \kappa } \sum_{k=n}^{\infty} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!} \end{eqnarray} $$ You are interested in the behavior of $2^n c_n(\ln 2)$ for large $n$.

The latter sum vanishes by the Stolz-Cesàro theorem. Indeed, for $b_n = n \kappa$, and $a_n = \sum \limits_{k=n}^{\infty} \frac{ \left(\ln\left( n \kappa\right)\right)^{k-1}}{(k-1)!}$, $$ \lim_{n \to \infty} \frac{a_{n+1} - a_{n}}{b_{n+1}-b_n} = \lim_{n \to \infty}\left(-\frac1{\kappa} \cdot \frac{ \left(\ln\left( n \kappa\right)\right)^{n-1}}{(n-1)!} \right)= 0 $$ where the Stirling approximation can be used to prove the limit.

Thus, $\lim \limits_{n \to \infty} c_n(\kappa) = 1$.

  • $\begingroup$ I especially appreciate the treatment of the iterated log, which was the sticking point for me. $\endgroup$
    – daniel
    Dec 22 '11 at 19:36

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