Let $d_n$ be the least common multiple of $1,2, \dots, n$.
Show that
$$\sum_{n=1}^{\infty} \frac{1}{d_n}$$
is an irrational number.
Scroll down for a solution.
Observation 1: If $n+1$ is a power of a prime $q$, then $d_{n+1} = q d_n$, otherwise $d_{n+1} = d_n$.
Observation 2:
$$\sum_{k=1}^{n} \frac{a_k - 1}{a_1 a_2 \dots a_k} = 1 - \frac{1}{a_1 a_2 \dots a_n}$$
(Proof left to reader).
Let the primes in order be $p_1, p_2, \dots$.
Pick an arbitrary prime $p = p_m$ and consider for $n \ge p$
$$f_n = \frac{d_{n}}{d_{p-1}}$$
The observation 1 above also holds for $f_n$.
Note that $f_{p_j} \ge p_m p_{m+1} \dots p_{j}$ and that the inequality is strict for infinitely many $p_j$.
Now consider $$ \sum_{p_i \le k < p_{i+1}} \frac{1}{f_k}$$
By Bertrands' theorem of a prime between $n$ and $2n$ we have that $p_{i+1} < 2p_i$
and thus
$$ \sum_{p_i \le k < p_{i+1}} \frac{1}{f_k} \le \frac{p_{i+1} - p_i}{f_{p_{i}}} \le \frac{p_i - 1}{f_{p_{i}}}$$
Since $f_{p_j} \ge p_m p_{m+1} \dots p_{j}$ we get
$$ \sum_{p_i \le k < p_{i+1}} \frac{1}{f_k} \le \frac{p_i - 1}{p_m p_{m+1} \dots p_i}$$
And so (note inequality is strict, because $f_{p_j} \gt p_m p_{m+1} \dots p_{j}$ infinitely often.)
$$\sum_{n \ge p} \frac{1}{f_n} < \sum_{j \ge m} \frac{p_j - 1}{p_m \dots p_j}$$
By Observation 2 above we get
$$\sum_{n \ge p} \frac{1}{f_n} < 1$$
Now if $$\frac{a}{b} = \sum_{n=1}^{\infty} \frac{1}{d_n}$$
Pick a prime $p > b$ and consider $\frac{a d_{p-1}}{b}$ and use the above result about $f$.
