At the first time I'm dealing with this problem:
prove that the series $\displaystyle\sum_{n=1}^{\infty} \frac{1}{d_{n}}$ is an irrational number, where $d_{n}=lcm(1,2,...,n)$.
After some calculations I'm reduced to prove that the series $\displaystyle\sum_{n=k+1}^{\infty} \frac{d_{k}}{d_{n}}$ tends to zero as n goes to infinity.
Now for this purpose I bounded this one with $\displaystyle \sum_{n=n_{0}}^{\infty} \frac{P_{n+1}-P_{n}}{\displaystyle\prod_{m=n_{0}}^{n} P_{m}}$, where $P_{m}$ is the $m-th$ prime, and $k+1=P_{n_{0}}$.
But it seems very difficult prove the result from this bound,because the presence of the gap between consecutive prime numbers.
So, can anyone help me to prove that the main series converge to zero using this methods or somelse? (I accept also some usefull hint).
Maybe I solved my problem..
we have to prove that the series write above can be taken arbitrary small when $n_{0}\rightarrow \infty$.
In fact I can write it as $ \frac{p_{n_{0}+1}}{p_{n_{0}}}-1 + \frac{p_{n_{0}+2}-p_{n_{0}+1}}{p_{n_{0}+1}}\frac{1}{p_{n_{0}}}+\frac{p_{n+1}-p_{n}}{p_{n}}\displaystyle \sum_{n=n_{0}+2}^{\infty} \frac{1}{\displaystyle\prod_{m=n_{0}}^{n-1} p_{m}}$.
Now the first term goes to $0$ since we know that $p_{n}\tilde{} nlog(n)$;
the second term goes to $0$ since $\frac{p_{n_{0}+2}-p_{n_{0}+1}}{p_{n_{0}+1}}<1$ by Bertrand`s postulate, and $\frac{1}{p_{n_{0}}}$ goes to $0$ when $n_{0}\rightarrow \infty$;
Finally the tird term goes to 0 since it is less then $ \displaystyle \sum_{n=n_{0}+2}^{\infty} \frac{1}{\displaystyle\prod_{m=n_{0}}^{n-1} p_{m}}\leq \displaystyle \sum_{n=n_{0}+2}^{\infty} \frac{1}{p_{n-1}p_{n-2}}$ and this goes to $0$ since the series $\sum_{n=3}^{\infty} \frac{1}{p_{n-1}p_{n-2}}$ converges because $p_{n}\tilde{} nlog(n)$.