素数乘积倒数和的级数紧界求解与收敛性问询

Absolutely yes! Let's walk through why: the general term of your series is (a_m = \frac{1}{p_1 p_2 \dots p_m}), where (p_1=2, p_2=3, p_3=5, \dots) are primes in order.

Primes grow at a rate of roughly (p_m \sim m \log m) (thanks to the Prime Number Theorem), so their product (called a primorial) blows up extremely fast—way quicker than any exponential function. As a quick sanity check: if we replace every prime after the first with 3 (a massive undercount of how fast primes grow), the terms become (\frac{1}{2 \cdot 3^{m-1}}), which is a convergent geometric series with sum (\frac{1/2}{1 - 1/3} = \frac{3}{4}).

Our original series has even smaller terms than this geometric series, so it definitely converges absolutely to a fixed constant. Calculating the first few terms gives a sense of its value:

• (a_1 = 1/2 = 0.5)
• (a_2 = 1/(2 \cdot 3) ≈ 0.1667)
• (a_3 = 1/(2 \cdot 3 \cdot 5) ≈ 0.0333)
• (a_4 = 1/(2 \cdot 3 \cdot 5 \cdot 7) ≈ 0.00476)
• (a_5 = 1/(2 \cdot 3 \cdot 5 \cdot 7 \cdot 11) ≈ 0.00043)

Adding these gets us to ~0.7052, and each subsequent term adds practically nothing (the 6th term is ~3.6e-5, the 7th ~3.0e-6, etc.). The total sum is a well-defined constant less than 1.

First, let's make sure we're on the same page: your series sums terms where each denominator is the product of primes up to (\sqrt{n}) (so the first (M) primes, where (p_M \le \sqrt{n}) and (p_{M+1} > \sqrt{n})).

Tight Bounds #

Since the infinite series converges to a constant ~0.7052, your partial sum (S(n)) is bounded above by this constant for all (n). For lower bounds:

• When (\sqrt{n} \ge 2) (i.e., (n \ge 4)), (S(n) \ge 1/2 = 0.5)
• When (\sqrt{n} \ge 3) (i.e., (n \ge 9)), (S(n) \ge 1/2 + 1/6 ≈ 0.6667)
• When (\sqrt{n} \ge 5) (i.e., (n \ge 25)), (S(n) \ge 0.6667 + 0.0333 = 0.7)

As (n) grows larger, (S(n)) gets closer and closer to that ~0.7052 value. So tight, practical bounds are:
[0.7 \le S(n) \le 0.706 \quad \text{for sufficiently large } n]
And for all (n), (0.5 \le S(n) < 1) is a simple, way tighter alternative to your existing (\sqrt{n}/2) bound.

Does (S(n) \le c\frac{1}{\sqrt{n}}) hold for some constant (c)? #

Nope, this can't be true. Here's the key reason: as (n) goes to infinity, (\sqrt{n}) grows without bound, so (c/\sqrt{n}) approaches 0. But (S(n)) approaches that fixed ~0.7052 constant. No matter what (c) you pick, eventually (n) will be large enough that (S(n)) is bigger than (c/\sqrt{n}) (since the left side stays near 0.7 and the right side shrinks to 0).

内容的提问来源于stack exchange，提问作者user437890