Mertens constant B_3

[Tomasz Ordowski]

Hello! 

Is it possible that 

B_3 = Sum_{n>=1} (log(p_n)/p_n - 1/n) ? 

Where B_3 = 1.33258227573322... 

See https://oeis.org/A083343

Or rather this series diverges to plus infinity?

Best, 

Tom Ordo  

______________

https://mathworld.wolfram.com/MertensConstant.html

See the last formula (18) for B_3.

[Gareth McCaughan]

Write A(x) = sum {p<x} of log p / p. The formula you cite says A(x) =
log x - B_3 + o(1).

Your sum truncated at n is A(p_n) - H_n = (log p_n - B_3 + o(1)) - (log
n + gamma + o(1)) = log (p_n/n) - (gamma+B_3) + o(1).

(H_n being the n'th harmonic number = 1 + 1/2 + ... + 1/n.)

p_n is about n log n, so log(p_n/n) is about log log n. So your series
diverges, albeit slowly. In particular: no, it is not possible that it
happens to converge to B_3.

--
g

[Tomasz Ordowski]

Yes, thanks! 

How about this?

Lim_{n->oo} (n / (p_n + n)) Sum_{k=1..n} (1/k + 1/p_k) = 1.

Best, 

Tom Ordo 

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[Tomasz Ordowski]

PS. I propose much stronger:

C = lim_{n->oo} (- p_n / n + Sum_{k=1..n} (1/k + 1/p_k)) > 2.  

Is this provable? 

Tom Ordo 

[Ruud H.G. van Tol]

On 2026-05-13 12:06, Tomasz Ordowski wrote:
> PS. I propose much stronger:
>
> C = lim_{n->oo} (- p_n / n + Sum_{k=1..n} (1/k + 1/p_k)) > 2.
>
> Is this provable?

To illustrate:

(PARI)
? t(n) = sum(k=1, n, 1/k + 1/prime(k)) - prime(n)/n;

? #[ n |n<-[1..1000], t(n) <= 2]
% = 347

   1..1000: 347
1001..2000:  77
2001..3000:  31
3001..4000:  13 (3462 is the last)
4001..5000:   0
5001..6000:   0
6001..7000:   0
...


So that is a sequence:

? [ n |n<-[1..72], t(n) <= 2]
% = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19,
20, 21, 22, 23, 24, 25, 26, 27, 28, 31, 32, 33, 34, 35, 36, 37, 38, 39,
40, 41, 42, 43, 44, 45, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58,
59, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72]

Skipped: [29, 30, 46, 60, 61, 90, ...]

-- Ruud

[Gareth McCaughan]

On 13/05/2026 09:46, Tomasz Ordowski wrote:
> Yes, thanks!
>
> How about this?
>
> Lim_{n->oo} (n / (p_n + n)) Sum_{k=1..n} (1/k + 1/p_k) = 1

The factor in front looks like 1 / log n. The two sums look like log n
and log log n, both with error terms that are o(log n), so yes, the
limit should be 1 unless I'm missing something.

--
g

[Gareth McCaughan]

On 13/05/2026 11:06, Tomasz Ordowski wrote:

PS. I propose much stronger:

C = lim_{n->oo} (- p_n / n + Sum_{k=1..n} (1/k + 1/p_k)) > 2.  

Is this provable?

This is a question about the error terms in a number of famous approximations, so let's take a look.

p_n / n = log n + log log n - 1 + o(1).

1+...+1/n = log n + gamma + o(1).

1/2+...+1/p_n = log log p_n + B + o(1) = log log n + B + o(1)

so this thing will tend to gamma + B + 1.

We have gamma ~= 0.577, B ~= 0.261, and 1 ~= 1.000 [citation needed], so it looks as if your limit is <2.

Ruud's calculations make it look otherwise. But here are some concrete values:

1 -0.5
10 1.5624070258402862
100 1.8837196391122415
1000 2.023882137261716
10000 2.0239642848417034
100000 1.9992015646171506
1000000 1.9750827709205847
10000000 1.959062951022723

-- 
g

[Tomasz Ordowski]

Yes, I agree with Gareth that C = 1 + gamma + B = 1.838712877749... 

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[Tomasz Ordowski]

PS. Could it be zero? 

Lim_{x->oo} (exp(-gamma)*Product_{p<=x} (1/(1-1/p)) - log x) = 0.  

https://en.wikipedia.org/wiki/Mertens%27_theorems#Changes_in_sign

[Tomasz Ordowski]

PS. Could it be zero? 

Lim_{x->oo} (exp(-gamma)*Product_{p<=x} (1/(1-1/p)) - log x) = 0.  

https://en.wikipedia.org/wiki/Mertens%27_theorems#Changes_in_sign

So, if this limit exists, it must be zero. But does it exist? 

Perhaps this follows from the Riemann Hypothesis. 

[Gareth McCaughan]

On 14/05/2026 16:23, Tomasz Ordowski wrote:
> PS. Could it be zero?
> Lim_{x->oo} (exp(-gamma)*Product_{p<=x} (1/(1-1/p)) - log x) = 0.
> https://en.wikipedia.org/wiki/Mertens%27_theorems#Changes_in_sign
> So, if this limit exists, it must be zero. But does it exist?

There's some discussion of this at
https://math.stackexchange.com/questions/3626427/error-term-in-mertens-third-theorem
where the first answer claims that

product (1-1/p) - exp(-gamma)/log x = O(exp(-c (log x)^a)) for any a<3/5
and a suitable choice of c (unconditionally)

and

product (1-1/p) - exp(gamma)/log x = O((log x)^2/sqrt(x)) on RH

and the same things should hold for your difference

exp(-gamma) product (1-1/p) - log x

so yes, it -> 0 provided that answer is correct.

The second answer gives a weaker unconditional bound but unlike the
first has a sketch of an actual proof. (I haven't attempted to check
whether it's correct.) This one too would imply that the limit is zero.

--
g

[Tomasz Ordowski]

Yes, thanks!
Similarly, we have (if I'm not mistaken): 

Lim_{x->oo} (exp(-gamma)*zeta(2)*Product_{p<=x} (1+1/p) - log x) = 0. 

Cf. https://mathworld.wolfram.com/MertensTheorem.html

(see last formulas). Best, Tom Ordo  

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