Saudi Arabia Mathematical Competitions 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 $$a_n\le 0.7-\frac{1}{4n},n\ge 3.(1)$$ 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 $$\begin{gathered} \frac{1}{2(2n+1)(n+1)}\le \frac{1}{4n}-\frac{1}{4(n+1)}\iff \\ \frac{1}{(n+1)(2n+1)}<\frac{1}{2n(n+1)}\iff \\ 2n<2n+1, \end{gathered}$$ and we are done.

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

---

Alternative solution.

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 $$b_n=1+\frac{1}{2}+\cdots +\frac{1}{n}-\ln n.$$ It is clear that $$b_{2n}-b_n=\frac{1}{n+1}+\cdots +\frac{1}{2n}-\ln 2=a_n-\ln 2.(1)$$ We will show that $(b_n{)}_{n\ge 1}$ is strictly decreasing. Indeed, we have $$\begin{gathered} b_{n+1}-b_n=\frac{1}{n+1}-\ln \frac{n+1}{n}=\frac{1}{n+1}\left[1-\ln {\left(1+\frac{1}{n}\right)}^{n+1}\right]<0, \\ \text{since }{\left(1+\frac{1}{n}\right)}^{n+1}>e,n=1,2,\dots \end{gathered}$$ 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.