$ \frac{1}{2\cdot 3}+\frac{1}{5\cdot 7}+\frac{1}{11\cdot 13}+\frac{1}{17\cdot 19}+\dots = \sum_{k=1}^{\infty}\frac{1}{p_{2k-1}\cdot p_{2k}}=16/75$?

Question

Basel problem solved by Euler is: $$ \sum_{k=1}^{\infty} \frac{1}{k^2}=\frac{\pi^2}{6} $$ Now , I want to know what is $$ \frac{1}{2\cdot 3}+\frac{1}{5\cdot 7}+\frac{1}{11\cdot 13}+\frac{1}{17\cdot 19}+\frac{1}{23\cdot 29}+ \dots = \sum_{k=1}^{\infty} \frac{1}{p_{2k-1}\cdot p_{2k}} == ? $$ where $p_k$ is the $k$'th prime. I tried writing a script

    <script async>
const primes_first_n10000=[null,2, 3, 5, 7, 11, 13,..., 104711, 104717, 104723, 104729];
let p =primes_first_n10000,n=800;
let a=1n,b=6n;
let t=6n;
for (let k=3;k<n;k+=2) {
    t=BigInt(p[k]*p[k+1])
    console.log(t)
    a=a*t+b
    b*=t;   
}
//console.log(a,'/',b/15n);
//  a/(b/15)===3.2===16/5 ?
console.log(a/16n,'---',b/75n);
    </script>

I guess that $\sum_{k=1}^{\infty} \frac{1}{p_{2k-1}\cdot p_{2k}}=\frac{16}{75}$ , however I'm not sure! Could you tell me the answer: what does the sum converge to?

#sagemath code
    var('k')
    #s=sum(1/k^2, k, 1, oo);print(s)
    s= sum(N(1/(nth_prime(k)*nth_prime(k+1)),100) for k in range(1,80000,2))
    print(N(s,100),N(pi/s,100))
    print(N(pi/15,100))
    print(N(s*15,100))

    Out: (0.21042571723113630717490968408, 14.929699159057624632476407160)
    0.20943951023931954923084289222
    3.1563857584670446076236452612

Being a user for over $9$ years , we expect you to properly format your question before posting. — The Demonix _ Hermit, Jan 10 '20 at 10:13
I'm a programmer,so I think some text like 1/(23)+1/(57)+1/(1113)+1/(1719)+1/(2329)+... == sum(1/(p[2k-1]p[2k]), k, 1, oo) is better than Latex! — a boy, Jan 10 '20 at 10:15
I really don't like Latex, MathJax ..., they are not math but symbol! — a boy, Jan 10 '20 at 10:16
The rendered view of MathJax/LaTeX code is very good, but latex code is bad, so I want to just render Sagemath code directly — a boy, Jan 10 '20 at 10:27
@aboy Remember that you are writing this question not for yourself , but for the community to read . Hence you should set aside your preference and try to write it in a way that most people are comfortable with . $($ Which in this case is $\LaTeX$ .$)$ — The Demonix _ Hermit, Jan 10 '20 at 10:34
See https://math.stackexchange.com/questions/2380962/how-to-approximately-compute-the-infinite-sum-of-inverses-of-products-of-consecu — lhf, Jan 10 '20 at 10:54
yes, /questions/2380962 asked the same series's value before two years ago, but he wrote as $ S = \sum_{i=1}^{\infty} \frac{1}{p_i p_{i+1}} $ . it's not clear. And, no one answered what does the sum converge to. — a boy, Jan 11 '20 at 14:46
I see no reason to believe that there is a closed-form expression for this sum. — Gerry Myerson, Jan 11 '20 at 16:04
No, /questions/2380962 doesn't ask about the same series. It asks about $$\frac{1}{2\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\frac{1}{7\cdot 11}+\frac{1}{11\cdot 13}+ \dots$$but you can use the same techniques shown there to bound your sum. — nickgard, Jan 11 '20 at 18:58

Oleg567 · Answer 1 · 2020-01-12T22:10:13.927

For $n>1$: $$p_n \ge 2n-1,$$ (strong inequality for $n>4$), so we can estimate upper bound for the series $$ S = \sum_{k=1}^{\infty}\dfrac{1}{p_{2k-1}p_{2k}}\tag{1} $$ this way: for some $K \in \mathbb{N}$: $$ S < \sum_{k=1}^{K}\dfrac{1}{p_{2k-1}p_{2k}} + \sum_{k=K+1}^{\infty}\dfrac{1}{(4k-3)(4k-1)} $$ $$ = \sum_{k=1}^{K}\dfrac{1}{p_{2k-1}p_{2k}} + \color{blue}{\sum_{k=1}^{\infty}\dfrac{1}{(4k-3)(4k-1)} } - \sum_{k=1}^{K}\dfrac{1}{(4k-3)(4k-1)} $$ $$ = \color{blue}{\dfrac{\pi}{8}} + \sum_{k=1}^{K}\dfrac{1}{p_{2k-1}p_{2k}} - \sum_{k=1}^{K}\dfrac{1}{(4k-3)(4k-1)}.\tag{2} $$

For $K=20$ we have: $$S < \color{blue}{0.392699...} + 0.209968... - 0.389574... \approx 0.213092...$$ which is (essentially) less than $16/75\approx0.213333...$ .

For $K=100$ we'll have better/closer upper bound: $S < 0.211000...$;
for $K=1000$ we have: $S < 0.210485...$;
for $K=10000$ we have: $S < 0.210431...$ .

Lower bound could be estimated simply via any partial sum: $$ S > \sum_{k=1}^{10000}\dfrac{1}{p_{2k-1}p_{2k}} \approx 0.210425... $$

So, finally: $$ 0.21042 < S < 0.21044. $$

(more precise value: $S = 0.21042575...$)

$ \frac{1}{2\cdot 3}+\frac{1}{5\cdot 7}+\frac{1}{11\cdot 13}+\frac{1}{17\cdot 19}+\dots = \sum_{k=1}^{\infty}\frac{1}{p_{2k-1}\cdot p_{2k}}=16/75$?

1 Answers1