Let $\left(a_{n}\right)$ be the sequence defined by $a_{1}=1$ and $a_{n+1}=1+a_{1} a_{2} \ldots a_{n}$. Determine $\sum_{n=1}^{+\infty} \frac{1}{a_{n}}$.
It is clear that for all $n \geqslant 2$, we have $a_{n+1}-1=a_{n}\left(a_{n}-1\right)$, and by inverting this relation, we get $\frac{1}{a_{n+1}-1}=\frac{1}{a_{n}-1}-\frac{1}{a_{n}}$, in other words $\frac{1}{a_{n}}=\frac{1}{a_{n}-1}-\frac{1}{a_{n+1}-1}$. By summing, we obtain $\sum_{k=1}^{n} \frac{1}{a_{k}}=1+\frac{1}{a_{2}-1}-\frac{1}{a_{n+1}-1}=2+\frac{1}{a_{n+1}-1}$. To conclude that $\sum_{n=1}^{+\infty} \frac{1}{a_{n}}=2$, it suffices to show that $\frac{1}{a_{n+1}-1} \underset{n \rightarrow+\infty}{ } 0$, in other words that $a_{n} \xrightarrow[n \rightarrow+\infty]{ }+\infty$. But this is clear because the definition of $\left(a_{n}\right)$ shows that $a_{n+1} \geqslant 1+a_{n}$, hence that $a_{n} \geqslant n$ for all $n$.

## 4 Test