Lemma 1.$n\geq 6$Let $p_{1}<\dots<p_{k}$ be all primes not greater than $n$ then $\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}<\frac{n-1}{2n}$ proof : \[\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}=\frac{1}{4}+\frac{1}{9}+\sum_{i=3}^{k}\frac{1}{p_{i}^{2}}< \frac{1}{4}+\frac{1}{9}+\sum_{l=5}^{n}\frac{1}{l^{2}}=\frac{1}{4}+\frac{1}{9}+\frac12\sum_{l=5}^{n}{(\frac{1}{l-1}-\frac{1}{l+1})}=\frac14+\frac19+\frac18-\frac{1}{n+1}<\frac{1}{2}-\frac{1}{2n}=\frac{n-1}{2n}\] Lemma 2:$a_{2}\dots a_{n}<\frac{1}{2^{n-1}}$ Proof:for n<$a_{2}\dots a_{n}\leq (\frac{a_{2}+a_{3}+\dots+a_{n}}{n-1})^{n-1}$ but $a_{2}+\dots+a_{n}\leq \sum_{i=1}^{k}{[\frac{n}{p_{i}}]\frac{1}{p_{i}}}\leq n\sum_{i=1}^{k}\frac{1}{p_{i}^{2}}\leq \frac{n-1}{2}$ (because of Lemma 1)thus $a_{2}\dots a_{n}<\frac{1}{2^{n-1}}\leq \frac{1}{2^{n-1}}$ with Lemma 1 and 2 we can easily prove the problem. I hope I am correct. I hope there is another proof.Anyone has another proof?Could somebody post it? Is my solution correct?Please tell me.

Hawk Tiger wrote: $\frac14+\frac19+\frac18-\frac{1}{n+1}$ This is not true... it has to be $\frac14+\frac19+\frac18-\frac{1}{2n+2}$ and then the inequality from Lemma 1 is true for $n>35,$ not for $n>5.$

The following equality holds: $\sum_{k=2}^na_k=\sum_{k=2}^n(\frac{1}{p_1}+\frac{1}{p_2}+\cdots+\frac{1}{p_k})=\sum_{p\leq n, p prime}\frac{1}{p}\left[\frac{n}{p}\right]$. The folowing inequalities are obvious for $n>0$: \[\sum_{p\leq n, p \text{prime}}\frac{1}{p} \left[\frac{n}{p}\right] \leq \sum_{p\leq n, \text{prime}}\frac{1}{p} \frac{n}{p}=n\sum_{p\leq n, \text{prime}} \frac{1}{p^2} \leq n(\frac14 +\sum_{k=1}^{\inf}\frac{1}{(2k+1)^2})< n(\frac14 +\frac14 \sum_{k=1}^{\inf}\frac{1}{k(k+1)})=\frac{n}{2}.\] Therefore $\sum_{k=2}^na_k<\frac{n}{2}$. By AM-GM $a_2a_3\cdots a_n<(\frac{a_2+\cdots+a_n}{n-1})^{n-1}<\frac{1}{2^{n-1}}(1+\frac{1}{n-1})^{n-1}<\frac{3}{2^n-1}$. Add this inequalities and obtain: $\sum_{n=2}^{\inf}a_2 \cdots a_n <\frac12+\frac16+\frac{1}{12}+\frac{1}{60}+3(\frac{1}{2^5}+\frac{1}{2^6}+\cdots)=\frac{229}{240}<1$.

My solution in near to Nazar's one, but I will post it anyway: Lemma 1: $\sum_{p\in P} \frac{1}{p^2}<\frac{1}{2}$. Proof: $\sum_{p\in P} \frac{1}{p^2}<\frac{1}{4}+\sum_{i=1}^{\infty}\frac{1}{(2i+1)^2}=\frac{1}{4}+\frac{\pi^2-8}{8}=\frac{\pi^2-6}{8}<\frac{10-6}{8}=\frac{1}{2}$. Lemma 2: $a_2+...+a_N<\frac{N}{2}$. Proof: for each $p_i, \frac{1}{p_i}$ is counted in our sum no more than $\frac{n}{p_i}$ times, so $a_2+...+a_N<N\cdot\sum_{p\in P} \frac{1}{p^2}<\frac{N}{2}$ due to lemma 1. Lemma 3: $a_2...a_N<\frac{3}{2^{N-1}}$. Proof: by AM-GM $a_2...a_N<\left (\frac{N}{2(N-1)}\right)^{N-1}<\frac{e}{2^{N-1}}<\frac{3}{2^{N-1}}$. Finally, $ \sum_{k=2}^{N} a_2a_3 \cdots a_k<\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{60}+\sum_{i=5}^{\infty} \frac{3}{2^i}<\frac{229}{240}$.