Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
the right hand side depends on $n$ :-) – vesszabo Aug 9 '12 at 20:58
up vote 5 down vote accepted

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

share|cite|improve this answer
I especially appreciate the treatment of the iterated log, which was the sticking point for me. – daniel Dec 22 '11 at 19:36

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.