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Looking at a theorem of Chebyshev, I noticed that $$\sum_{n=0}^{\infty} \sum_{p_k < n} \frac{(\log p_k)^n}{n!} = 2 + 3 + ...+ p_k.$$

Proof. Letting $x = \log p_k$ and writing out the expansion of $e^{x}$ we get

$\lim\limits_{n\to\infty}\bigl[1 + \log 2 + \frac{\log (2)^2}{2} + \frac{\log(2)^3}{3!} + ... + \frac{\log(2)^n}{n!} \bigr]= 2;$

$\lim\limits_{n\to\infty}\bigl[1 + \log 3 + \frac{\log(3)^2}{2} + \frac{\log(3)^3}{3!} + ... + \frac{\log(3)^n}{n!} \bigr] = 3;$

$\lim\limits_{n\to\infty}\bigl[1 + \log 5 + \frac{\log(5)^2}{2} + \frac{\log(5)^3}{3!} +...+ \frac{\log(5)^n}{n!} \bigr] = 5;$

$\hspace{5cm}\vdots$

$\lim\limits_{n\to\infty}\bigl[1 + \log p_k + \frac{\log(p_k)^2}{2} + \frac{\log(p_k)^3}{3!} +...+ \frac{\log(p_k)^n}{n!} \bigr] = p_k $

The sum (from columns) is

$$S = k + \sum \log p_k + \sum (\log p_k)^2/2 + \sum (\log p_k)^3/6 +... + \sum (\log p_k)^n/n! = \sum p_k.$$

My question is: whether the above is correct?

Mhenni Benghorbal
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daniel
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  • Yes, but what does this have to do with number theory or primes? It is true for any finite sequence of real numbers $p_k > 0$. – Erick Wong Jul 13 '13 at 06:30
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    @Erick Wong:As it stands you are right. The original post had an additional note about a connection with the PNT. If I can fix my question I will re-post separately and maybe link to this. But maybe should be re-tagged. – daniel Jul 13 '13 at 11:32

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Yes, your reasoning is correct: recall that if $(A_n), (B_n),\dots, (Z_n)$ are convergent sequences, then $$\lim_{n\to\infty}(A_n+B_n+\cdots+Z_n)=\Bigl(\lim_{n\to\infty}A_n\Bigl)+\Bigl(\lim_{n\to\infty}B_n\Bigl)+\cdots+\Bigl(\lim_{n\to\infty}Z_n\Bigl)\,,$$

and don't forget that series are sequences of partial sums (note: the result above is also valid for more than 26 sequences ;-) )

Matemáticos Chibchas
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