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Let $a_n$ be a strictly increasing sequence of positive integers. Prove that

$$x_n =\frac{1}{a_1} + \frac{1}{a_1 a_2} +\cdots + \frac{1}{a_1 a_2 a_3 \cdots a_n}$$

converges to an irrational number.

Solution


Since $a_n \geq n$ holds , then:

$$\frac{1}{{{a_1}{a_2} \cdots {a_n}}} \le \frac{1}{{1 \cdot 2 \cdots n}} = \frac{1}{{n!}}$$

for each positive integer $n$ . Hence from the comparison test the series $\displaystyle \sum\limits_{n = 1}^\infty  {\frac{1}{{{a_1}{a_2} \cdots {a_n}}}}$ converges to a real number, call that $x$.

Suppose that $x$ is rational, that is $x=p/q$. Then there exists a positive integer $k$ such that:

$${a_k} \le q < {a_{k + 1}}$$

We consider the number $\displaystyle y = q{a_1}{a_2} \cdots {a_k}\left( {x - \sum\limits_{n = 1}^k {\frac{1}{{{a_1}{a_2} \cdots {a_n}}}} } \right)$ and we observe that

$$y = p{a_1}{a_2} \cdots {a_k} -q \sum\limits_{n = 1}^k {\frac{{{a_1}{a_2} \cdots {a_k}}}{{{a_1}{a_2} \cdots {a_n}}}}$$

hence $y$ is an integer.

Also it holds that:

$$\begin{aligned}
0&< y = q{a_1}{a_2} \cdots {a_k}\sum\limits_{n = k + 1}^\infty  {\frac{1}{{{a_1}{a_2} \cdots {a_n}}}}\\
 &=q\sum\limits_{n = k + 1}^\infty  {\frac{1}{{{a_{k + 1}}{a_{k + 2}} \cdots {a_n}}}} \\
 &\leq q\sum\limits_{n = k + 1}^\infty  {\frac{1}{{\left( {q + 1} \right)\left( {q + 2} \right) \cdots \left( {q + n - k} \right)}}} \\
 &< q\sum\limits_{n = k + 1}^\infty  {\frac{1}{{{{\left( {q + 1} \right)}^{n - k}}}}} \\
 &=q\sum\limits_{n = 1}^\infty  {\frac{1}{{{{\left( {q + 1} \right)}^n}}}}  = q\frac{{\frac{1}{{q + 1}}}}{{1 - \frac{1}{{q + 1}}}} \\
 &=1
\end{aligned}$$

leading us a to contradiction. Hence $x$ is irrational.