$(an)_{n\in\mathbb{N}^*}$ is a sequence, so that $a_1=1$, for all $n\geq 2$, $a_n=a_1\cdot...\cdot a_{n-1}+1$. $\sum_{n=1}^{m}\frac{1}{a_n}<M$, find M

Question

This is a problem I came across on another exam. It is as follows:

Let $(an)_{n\in\mathbb{N}^*}$ a sequence such that $a_1=1$ and for all $n\geq 2$, $a_n=a_1\cdot...\cdot a_{n-1}+1$. Find the real number $M$ such that $\sum_{n=1}^{m}\frac{1}{a_n}<M$.

Apparently, $M$ has a "nice" value.

I begin by noticing a pattern in the sequence.

$$a_1=1$$

$$a_2=a_1+1=2$$

$$a_3=a_1\cdot a_2+1=3$$

$$a_4=a_1 \cdot a_2 \cdot a_3 +1=7$$

$$a_5=a_1 \cdot a_2 \cdot a_3 \cdot a_4 +1=43$$

$$....$$

We can also observe that:

$$\frac{a_{n+1}}{a_n}=\frac{a_1 \cdot a_2 ...\cdot a_n +1}{a_n}$$

$$\frac{a_1 \cdot a_2 ...\cdot a_n}{a_n}+\frac{1}{a_n}$$

$$\frac{(a_n)^2-a_n}{a_n}+\frac{1}{a_n}$$

$$\frac{a_{n+1}}{a_n}=a_n-1 + \frac{1}{a_n}$$

However this doesn't really seem too helpful.

I also attempted to substitute the values directly into the summation:

$$\sum_{n=1}^{m}\frac{1}{a_n}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{7}...+\frac{1}{a_m}$$

However I'm not really sure how to proceed further and arrive at a "nice" value for M. Help would be appreciated!

Answer 1

(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 $m$.