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How to find these limits
$\displaystyle\lim_{n\to\infty}\left(\ln(\ln(n)) - \sum_{k=2}^n\frac1{k \ln(k)}\right)$ ?

and $\displaystyle\lim_{n\to\infty}\left( \ln(\ln(n)) - \sum_{k=1}^n\frac1{p_k}\right)$?

where $p_k$ is the $k$'th prime number.

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slightly related to… – TROLLKILLER Nov 5 '11 at 23:48
Would a rational expression whose denominator was kPi(k) be defined if Pi(k) could equal o? Perhaps I am missing something? – daniel Nov 5 '11 at 23:53
@daniel: oh, zero is a prime now? – J. M. Nov 6 '11 at 0:22
@J.M. The question was edited about the time I posted my comment. It originally read, "...where Pi(k) = 0 if k is not prime." It still needs editing. It now reads "where Pi(k) the k'th prime number." – daniel Nov 6 '11 at 0:36
-1 if possible to J.M. for sarcastic remark against a good comment – Graphth Nov 6 '11 at 0:43
up vote 3 down vote accepted

The second limit is precisely Mertens Constant.

The constant of the first limit, lets call it $C_{-1}$. I am not sure if it has a name. I believe Ramanujan computed that it was approximately $\approx 0.7946786$. See page 11 of this PDF for more details.

Remarkably it also appears in the following limit due to Ramanujan:

$$\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=2}^\infty \frac{1}{k(k^{\frac{1}{n}}-1)}-\log n=C_{-1}.$$

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In Mertens' constant, the sum is over the primes not exceeding $n$. In the current question, the sum is over the first $n$ primes (if we have figured out what OP wanted to say). It is not clear to me that the two different sums yield the same constant. – Gerry Myerson Nov 7 '11 at 1:03
@GerryMyerson: I didn't see that, but it is still the same constant. Since $p_k\sim k\log k$ we are looking at $$\log\log \left(k\log k+o(k\log k)\right)+B_1+o(1).$$ If we carefully work the error through the logarithms, it remains $$\log \log k+B_1+o(1).$$ – Eric Naslund Nov 7 '11 at 1:08
For the 1st limit, the author just gives that number, and a citation of a paper of Boas. But Boas gives the constant as .42816572, so I don't know where the author gets .7946786. – Gerry Myerson Nov 7 '11 at 1:17
@GerryMyerson: I am pretty sure that 0.7946786 is correct. I computed it numerically, and then googled the number I found. That is how I found the references to Ramanujans Notebooks in the first place, and the other paper. I am not sure if google books links work, but try:… – Eric Naslund Nov 7 '11 at 2:30
I think what I missed is that Boas is talking about $\int_2^n(dx/x\log x)$ while we're talking about $\log\log n$, and they differ by $\log\log2$. – Gerry Myerson Nov 7 '11 at 3:12

Mertens proved the existence of $$\lim(\sum_{p\le n}(1/p)-\log\log n)$$ see here for more detail, or any good textbook for a proof. This isn't quite what's wanted in the second problem above, but it should get you started.

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