(This is essentially taken from Sum of inverses of a sequence on AoPS.)

For all $n \ge 2$ is $$ \frac{1}{a_1 \cdots a_{n-1}} = \frac{a_n}{a_1 \cdots a_n } = \frac{a_1\cdots a_{n-1} + 1}{a_1 \cdots a_n} = \frac{1}{a_n} + \frac{1}{a_1 \cdots a_{n}} \, , $$ i.e. $$ \frac{1}{a_n} = \frac{1}{a_1 \cdots a_{n-1}} - \frac{1}{a_1 \cdots a_{n}} \, . $$ It follows that $$ \sum_{n=1}^m \frac{1}{a_n} = \frac{1}{a_1} + \sum_{n=2}^m \left( \frac{1}{a_1 \cdots a_{n-1}} - \frac{1}{a_1 \cdots a_{n}}\right) = \frac{2}{a_1} - \frac{1}{a_1 \cdots a_m} $$ for all $m \ge 1$, note that the sum is a “telescoping sum.” So we have $$ \sum_{n=1}^m a_n = 2 - \frac{1}{a_1 \cdots a_m} < 2 $$ for all $m \ge 2$, which proves that $M=1$ is an upper bound for $\sum_{n=1}^m a_n$. That is also the smallest upper bound, since $a_1 \cdots a_m \to \infty$ and therefore $\frac{1}{a_1 \cdots a_m} \to 0$ for $m \to \infty$.

In other words: $M = \sum_{n=1}^\infty a_n = 2$.

(This is essentially taken from Sum of inverses of a sequence on AoPS.)

For all $n \ge 2$ is $$ \frac{1}{a_1 \cdots a_{n-1}} = \frac{a_n}{a_1 \cdots a_n } = \frac{a_1\cdots a_{n-1} + 1}{a_1 \cdots a_n} = \frac{1}{a_n} + \frac{1}{a_1 \cdots a_{n}} \, , $$ i.e. $$ \frac{1}{a_n} = \frac{1}{a_1 \cdots a_{n-1}} - \frac{1}{a_1 \cdots a_{n}} \, . $$ It follows that $$ \sum_{n=1}^m \frac{1}{a_n} = \frac{1}{a_1} + \sum_{n=2}^m \left( \frac{1}{a_1 \cdots a_{n-1}} - \frac{1}{a_1 \cdots a_{n}}\right) = \frac{2}{a_1} - \frac{1}{a_1 \cdots a_m} $$ for all $m \ge 1$, note that the sum is a “telescoping sum.” So we have $$ \sum_{n=1}^m \frac{1}{a_n} = 2 - \frac{1}{a_1 \cdots a_m} < 2 $$ for all $m$, which proves that $M=2$ is an upper bound for $\sum_{n=1}^m \frac{1}{a_n}$. That is also the smallest upper bound, since $a_1 \cdots a_m \to \infty$ and therefore $\frac{1}{a_1 \cdots a_m} \to 0$ for $m \to \infty$.

In other words: $M = \sum_{n=1}^\infty \frac{1}{a_n} = 2$ is the smallest number with the property that $\sum_{n=1}^{m}\frac{1}{a_n}<M$ for all $m$.

(This is essentially taken from Sum of inverses of a sequence on AoPS.)

For all $n \ge 2$ is $$ \frac{1}{a_1 \cdots a_{n-1}} = \frac{a_n}{a_1 \cdots a_n } = \frac{a_1\cdots a_{n-1} + 1}{a_1 \cdots a_n} = \frac{1}{a_n} + \frac{1}{a_1 \cdots a_{n}} \, , $$ i.e. $$ \frac{1}{a_n} = \frac{1}{a_1 \cdots a_{n-1}} - \frac{1}{a_1 \cdots a_{n}} \, . $$ It follows that $$ \sum_{n=1}^m \frac{1}{a_n} = \frac{1}{a_1} + \sum_{n=2}^m \left( \frac{1}{a_1 \cdots a_{n-1}} - \frac{1}{a_1 \cdots a_{n}}\right) = \frac{2}{a_1} - \frac{1}{a_1 \cdots a_m} $$ for $m \ge 2$, note that the sum is a “telescoping sum.” So we have $$ \sum_{n=1}^m a_n = 2 - \frac{1}{a_1 \cdots a_m} < 2 $$ for all $m \ge 2$, which proves that $M=1$ is an upper bound for $\sum_{n=1}^m a_n$. That is also the smallest upper bound, since $a_1 \cdots a_m \to \infty$ and therefore $\frac{1}{a_1 \cdots a_m} \to 0$ for $m \to \infty$.

In other words: $M = \sum_{n=1}^\infty a_n = 2$.

(This is essentially taken from Sum of inverses of a sequence on AoPS.)

For all $n \ge 2$ is $$ \frac{1}{a_1 \cdots a_{n-1}} = \frac{a_n}{a_1 \cdots a_n } = \frac{a_1\cdots a_{n-1} + 1}{a_1 \cdots a_n} = \frac{1}{a_n} + \frac{1}{a_1 \cdots a_{n}} \, , $$ i.e. $$ \frac{1}{a_n} = \frac{1}{a_1 \cdots a_{n-1}} - \frac{1}{a_1 \cdots a_{n}} \, . $$ It follows that $$ \sum_{n=1}^m \frac{1}{a_n} = \frac{1}{a_1} + \sum_{n=2}^m \left( \frac{1}{a_1 \cdots a_{n-1}} - \frac{1}{a_1 \cdots a_{n}}\right) = \frac{2}{a_1} - \frac{1}{a_1 \cdots a_m} $$ for all $m \ge 1$, note that the sum is a “telescoping sum.” So we have $$ \sum_{n=1}^m a_n = 2 - \frac{1}{a_1 \cdots a_m} < 2 $$ for all $m \ge 2$, which proves that $M=1$ is an upper bound for $\sum_{n=1}^m a_n$. That is also the smallest upper bound, since $a_1 \cdots a_m \to \infty$ and therefore $\frac{1}{a_1 \cdots a_m} \to 0$ for $m \to \infty$.

In other words: $M = \sum_{n=1}^\infty a_n = 2$.