Let $a_1,a_2,a_3,\ldots,a_{250}$ be real numbers such that $a_1=2$ and $$a_{n+1}=a_n+\frac{1}{a_n^2}$$ for every $n=1,2, \ldots, 249$. Let $x$ be the greatest integer which is less than $$\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_{250}}$$ How many digits does $x$ have? Proposed by Miroslav Marinov, Bulgaria
Problem
Source: JBMO Shortlist 2023, A7
Tags: Sequence, inequalities, JBMO, JBMO Shortlist, algebra
pco
28.06.2024 11:08
Orestis_Lignos wrote: Let $a_1,a_2,a_3,\ldots,a_{250}$ be real numbers such that $a_1=2$ and $$a_{n+1}=a_n+\frac{1}{a_n^2}$$ for every $n=1,2, \ldots, 249$. Let $x$ be the greatest integer which is less than $$\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac{1}{a_{250}}$$ How many digits does $x$ have? Let $S=\sum_{i=1}^{250}\frac 1{a_i}$ 1) $S>10$ $a_n$ is increasing and so $a_n\ge 2$ $\forall n$ and so $a_{n+1}\le a_n+\frac 14$ and so $a_n\le 2+\frac{n-1}4$ $=\frac {n+7}4$ So $S\ge\sum_{n=1}^{250}\frac 4{n+7}\ge\int_1^{251}\frac {4dx}{x+7}$ $=4(\ln 258 -\ln 8)=4\ln\frac{258}8>4\log_432=4\times\frac 52=10$ 2) $S<50$ $a_{n+1}^3=a_n^3+3+\frac 3{a_n^3}+\frac 1{a_n^6}>a_n^3+3$ and so $a_n^3\ge 2^3+3(n-1)=3n+5$ And so $a_n\ge\sqrt[3]{3n+5}$ So $S\le \sum_{n=1}^{250}\frac 1{\sqrt[3]{3n+5}}<\int_0^{250}\frac{dx}{\sqrt[3]{3x+5}}$ $=\frac 12\left[(3x+5)^{\frac 23}\right]_0^{250}$ $=\frac{755^{\frac 23}-5^{\frac 23}}2<\frac{1000^{\frac 23}}2=50$ Hence the answer $\boxed{\lfloor S\rfloor\text{ has two decimal digits}}$
Assassino9931
02.07.2024 00:03
My problem. In the current version one does not need super tight estimates with integrals and that is why it is suitable for juniors, reasonably crude estimates and thinking work.
Denote $\displaystyle S = \frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_{250}}$, it suffices to show \[10 < S < 100.\]
We compute $a_1 = 2$, $a_2 = \frac{9}{4}$, $a_3 = \frac{793}{324} \in \left(\frac{12}{5}, \frac{5}{2}\right)$ and so $a_4 = a_3 + \frac{1}{a_3^2} > \frac{12}{5} + \frac{4}{25} = \frac{64}{25}$. All terms are positive, so $a_{n+1} > a_n$ for $n\geq 1$; in particular, $a_n \geq a_4 > \frac{64}{25}$ for $n=4,5,\ldots,250$. Therefore
\[S < \frac{1}{2} + \frac{4}{9} + \frac{2}{5} + 247 \cdot \frac{25}{64} < 2 + 98 = 100.\]
On the other hand, $a_n > a_2 = \frac{9}{4}$ for $n\geq 2$, we have $a_{n+1} < a_n + \frac{16}{81} < a_n + \frac{1}{5}$ for $n\geq 2$.Hence (by induction) $a_n < \frac{n+10}{5}$ for all $n\geq 2$, so
\[ S > \frac{1}{2} + 5 \cdot \left(\frac{1}{12} + \frac{1}{13} + \frac{1}{14} + \cdots + \frac{1}{260}\right).\]The latter expression contains $8$ fractions over $\frac{1}{20}$, $30$ other fractions over $\frac{1}{50}$, $50$ other fractions over $\frac{1}{100}$ and $100$ other fractions over $\frac{1}{200}$. Therefore the expression exceeds $5\left(\frac{8}{20} + \frac{3}{5} + \frac{1}{2} + \frac{1}{2}\right) = 10$. In conclusion, $S > 10$.
Thanks to @dangerousliri and @Marinchoo for discussing the following. Direct verification shows that $x + \frac{1}{x^2}$ increases for $x\geq 2$, hence one may show (by induction) a significantly better bound for $a_n$, say. $\sqrt[3]{3n+5} \leq a_n \leq \sqrt[3]{4n+4}$ for all $n$. In other words, $a_n$ grows at rate of $n^{1/3}$. A more direct way to see this is as follows: $a_{n+1}^3 = a_n^3 + 3 + \frac{3}{a_n^3} + \frac{1}{a_n^8}$, so $a_{n+1}^3 - a_n^3 \in [3,4]$ for all $n$, whence (by induction) $\sqrt[3]{3n+5} \leq a_n \leq \sqrt[3]{4n+4}$. In a similar fashion one obtains that for all $\varepsilon > 0$ and sufficiently large $n$ one has $\sqrt[3]{3n} < a_n < \sqrt[3]{(3+\varepsilon)n}$.