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My question is why the following happens, and whether we can correct (2) below to account for an errant factor of 2.

By a slight generalization* of the argument of this problem we have I think that

$$\sum_{p<x}(\log p)^m \sim (\log x)^m\cdot \pi(x).\hspace{20mm}(1)$$

And from an earlier question we have that

$$\sum_{p_k < x} p_k = \lim_{n\to \infty}\left( k + \sum \log p_k + \sum (\log p_k)^2/2 +...+ \sum (\log p_k)^n/n!\right)$$

So I thought to re-write the last in terms of (1):

$$\sum p_k = \lim_{n\to\infty} \pi(x) \left(1 + \log x + (\log x)^2/2 +...+(\log x)^n/n! \right) \hspace{20mm}(2)$$

But the right-hand side is too big by a factor of 2 (very nearly).

Of course I went back and looked at (1) but for each m it is numerically plausible. When I compare the sums I see there are numerical discrepancies that accumulate in (2) without affecting (1). Is there some aspect of the error of (1) that would give such an exact discrepancy in (2)?

Thanks for any suggestions.

Edit--As a quick check, (2) (with the correction of division by 2 on the r.h.s.) would imply $\sum_{p < x} p \sim \frac{x^2}{2\log x}$ which is at least numerically plausible.

*I can post the proof as a separate question if it is in doubt.

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daniel
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1 Answers1

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With $p_k \sim k\log k$, you indeed have $\sum\limits_{p < x} p \sim \frac{x^2}{2\log x}$.

The problem in your formula $(2)$ is that you can't blithely replace expressions with asymptotically equal expressions in a sum, when the number of terms in the sum tends toward infinity.

When we compute $\sum\limits_{p < x} (\log p)^m$, let us not forget everything below the highest-order term.

$$\begin{align} \sum_{p < x}(\log p)^m &= \sum_{k < x}(\log k)^m\bigl(\pi(k) - \pi(k-1)\bigr)\\ &= \sum_{k < x}\pi(k)(\log k)^m - \sum_{k < x-1}\pi(k)\bigl(\log (k+1)\bigr)^m\\ &\approx \pi(x)(\log x)^m - \sum_{k < x-1}\pi(k)\left(\bigl(\log(k+1)\bigr)^m - (\log k)^m\right)\\ &\approx \pi(x)(\log x)^m - m\sum_{k < x-1}\pi(k)(\log k)^{m-1}\log(1 + 1/k)\\ &\approx \pi(x)(\log x)^m - m\sum_{k < x-1}\frac{\pi(k)}{k}(\log k)^{m-1}\\ &\approx \pi(x)(\log x)^m - m\sum_{k < x-1}(\log k)^{m-2} \end{align}$$

A crude estimate for $\sum\limits_{k < y}(\log k)^r$ yields

$$\frac{y}{2}(\log y)^r\left(1 - \frac{\log 2}{\log y}\right)^r \leqslant \sum_{k < y}(\log k)^r \leqslant y(\log y)^r,$$

and for the sum of the $m$-th power of the prime logarithms, we get

$$\sum_{p < x}(\log p)^m \approx \pi(x)(\log x)^m - cmx(\log x)^{m-2} \approx \pi(x)(\log x)^m\left(1 - \frac{cm}{\log x}\right)$$

where $c$ is not much smaller than $\frac12$.

If we insert that approximation int $(2)$, the general term in the sum becomes

$$\frac{(\log x)^m}{m!}\left(1 - \frac{cm}{\log x}\right) = \frac{(\log x)^m}{m!} - c\frac{(\log x)^{m-1}}{(m-1)!}$$

and, with some handwaving can attribute the spurious factor $2$ to the correction terms.

(Please don't ask me to make the estimates precise ;)

Daniel Fischer
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  • I had to work at the first equivalence a bit but I see that by letting the upper limit for the sum $\sum k\log k$ be $\pi(x)$ the result follows. Thanks. – daniel Jul 15 '13 at 03:37