2
$\begingroup$

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.

  • 0
    slightly related to http://math.stackexchange.com/questions/79115/limit-lim-limits-n-rightarrow-infty-left2-sqrt-n-sum-limits-k-1n-frac2011-11-05

2 Answers 2

3

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:

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

  • 0
    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.2011-11-07
  • 0
    @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 $$\log \log k+B_1+o(1).$$2011-11-07
  • 0
    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 .7946786.2011-11-07
  • 0
    @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: http://books.google.ca/books?id=liTRR8UnTq4C&pg=PA166&lpg=PA166&dq=0.7946786&source=bl&ots=mbTIgrkyRL&sig=TBkzDvLKJLUEGq8xeszGPfBPano&hl=en&ei=4UK3TuOrLcnN4QTPtN3zAw&sa=X&oi=book_result&ct=result&resnum=1&ved=0CBsQ6AEwAA#v=onepage&q=0.7946786&f=false2011-11-07
  • 0
    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 $\log\log2$.2011-11-07
1

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.