How to approximately compute the infinite sum of inverses of products of consecutive primes

Question

Let $$S = \sum_{i=1}^{\infty} \frac{1}{p_i p_{i+1}},$$ where $p_i$ denotes the $i$-th prime number.

How to approximately calculate $S$? Is there a way of quickly calculating the first one/two/three digits of $S$, without resorting to computer programs? (but using calculators is allowed)

EDIT: replaced notation $\pi(i)$ with $p_i$. Added note that using calculators is allowed, but computer programs not.

Answer 1

Some observations: the standard notation for the $k$-th prime is $p_{k}$ or $q_{k}$. The symbol $\pi\left(k\right)$ usually represents the counting function of primes up to $k$, i.e., $$\pi\left(k\right)=\sum_{p\leq k}1,\,p\mathrm{\,is\,a\,prime\,number}.$$ Assuming that with the symbol $\pi\left(k\right)$ you intend $p_{k}$, you may observe that $$\sum_{k\geq n}\frac{1}{p_{k}p_{k+1}}\leq\sum_{k\geq n}\frac{p_{k+1}-p_{k}}{p_{k}p_{k+1}}=\sum_{k\geq n}\left(\frac{1}{p_{k}}-\frac{1}{p_{k+1}}\right)=\frac{1}{p_{n}}$$ and since we know that $$p_{n}>n\log\left(n\log\left(n\right)\right)-n,\,n\geq2$$ we get $$S<\sum_{k\leq n-1}\frac{1}{p_{k}p_{k+1}}+\frac{1}{n\log\left(n\log\left(n\right)\right)-n}$$ so taking, for example, $n=11$, we obtain $$S<\sum_{k\leq10}\frac{1}{p_{k}p_{k+1}}+\frac{1}{11\log\left(11\log\left(11\right)\right)-11}\approx \color{red}{0.334}.$$

Answer 2

You can compute a few terms and then bound the remaining terms by $$ \sum_{i=n}^{\infty} \frac{1}{\pi(i) \pi(i+1)} \le \sum_{i=n}^{\infty} \frac{1}{i(i+1)} = \sum_{i=n}^{\infty} \left( \frac{1}{i} - \frac{1}{i+1} \right) = \frac{1}{n}. $$