, 2012

The numbers $a_1 = \frac{1}{2}$ and $a_2 = \frac{7}{12}$ have the first decimal 5. The number
$$a_3 = \frac{1}{4} + \frac{1}{5} + \frac{1}{6} = \frac{37}{60} > 0.6$$
has the first decimal 6.

We will prove that for every $n \ge 3$, we have $0.6 < a_n < 0.7$. Notice that
$$a_{n+1} - a_n = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)} > 0,$$
so $a_n \ge a_3 > 0.6$, for every $n \ge 3$.

We will prove by induction that
$$a_n \le 0.7 - \frac{1}{4n}, \quad n \ge 3. \quad (1)$$
For $n=3$, we have $a_3 = \frac{37}{60} = 0.7 - \frac{1}{12}$. Assume that
$$a_n \le 0.7 - \frac{1}{4n}.$$
Then we get
$$a_{n+1} = a_n + \frac{1}{(2n+1)(2n+2)} \le 0.7 - \frac{1}{4n} + \frac{1}{(2n+1)(2n+2)} < 0.7 - \frac{1}{4(n+1)},$$
since we have
$$\frac{1}{2(2n+1)(n+1)} \le \frac{1}{4n} - \frac{1}{4(n+1)} \Leftrightarrow \\ \frac{1}{(n+1)(2n+1)} < \frac{1}{2n(n+1)} \Leftrightarrow \\ 2n < 2n+1,$$
and we are done.

For $n \ge 3$, the first decimal of $a_n$ is 6.

We have
$$a_{n+1} - a_n = \frac{1}{2n+1} + \frac{1}{2n+2} - \frac{1}{n+1} = \frac{1}{(2n+1)(2n+2)} > 0,$$
so the sequence $(a_n)_{n \ge 1}$ is strictly increasing.

Consider the sequence $(b_n)_{n \ge 1}$, defined by
$$b_n = 1 + \frac{1}{2} + \dots + \frac{1}{n} - \ln n.$$
It is clear that
$$b_{2n} - b_n = \frac{1}{n+1} + \dots + \frac{1}{2n} - \ln 2 = a_n - \ln 2. \quad (1)$$
We will show that $(b_n)_{n \ge 1}$ is strictly decreasing. Indeed, we have
$$b_{n+1} - b_n = \frac{1}{n+1} - \ln \frac{n+1}{n} = \frac{1}{n+1} \left[ 1 - \ln \left( 1 + \frac{1}{n} \right)^{n+1} \right] < 0, \\ \text{since } \left(1 + \frac{1}{n}\right)^{n+1} > e, n = 1, 2, \dots$$
It follows that $b_{2n} - b_n > 0$, and hence from (1) we obtain $a_n < \ln 2 < 0.7$. As in the previous solution, for $n \ge 3$, we have $0.6 \le a_n < 0.7$, so hence the first decimal in this case is 6.