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

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. Jan 10, 2020 at 10:13
• But the rest of Math.stackexchange likes MathJax/LaTeX... Jan 10, 2020 at 10:18
• Jan 10, 2020 at 10:21
• @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 $\LaTeX$ .$)$ Jan 10, 2020 at 10:34
• – lhf
Jan 10, 2020 at 10:54

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: $$S = 0.21042575...$$)