# 1970 IMO Problems/Problem 3

## Problem

The real numbers $a_0, a_1, \ldots, a_n, \ldots$ satisfy the condition: $1 = a_{0} \leq a_{1} \leq \cdots \leq a_{n} \leq \cdots$.

The numbers $b_{1}, b_{2}, \ldots, b_n, \ldots$ are defined by $b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}}$

(a) Prove that $0 \leq b_n < 2$ for all $n$.

(b) given $c$ with $0 \leq c < 2$, prove that there exist numbers $a_0, a_1, \ldots$ with the above properties such that $b_n > c$ for large enough $n$.

## Solution $b_n = \sum_{k=1}^{n} \frac{1 - \frac{a_{k-1}}{a_{k}} }{\sqrt{a_k}} = \sum_{k=1}^{n} \frac{a_k - a_{k-1}}{a_k\sqrt{a_k}} = \sum_{k=1}^{n} (a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)$

Let $X_k$ be the rectangle with the verticies: $(a_{k-1},0)$; $(a_{k},0)$; $(a_{k},a_k^{-\dfrac{3}{2}})$; $(a_{k-1},a_k^{-\dfrac{3}{2}})$. $[asy] import graph; size(10cm,10cm,IgnoreAspect); Label f; f.p=fontsize(6); xaxis(0,10); yaxis(0,1); real f(real x) { return x^(-3/2); } draw(graph(f,1,10)); draw((1,0)--(1,1)); draw((1,0)--(2,0)--(2,2^(-3/2))--(1,2^(-3/2))--cycle); draw((2,0)--(3,0)--(3,3^(-3/2))--(2,3^(-3/2))--cycle); draw((3,0)--(4,0)--(4,4^(-3/2))--(3,4^(-3/2))--cycle); draw((6,0)--(7,0)--(7,7^(-3/2))--(6,7^(-3/2))--cycle); draw((7,0)--(8,0)--(8,8^(-3/2))--(7,8^(-3/2))--cycle); label("X_1",(1.5,0),0.5*N); label("X_2",(2.5,0),0.5*N); label("X_3",(3.5,0),0.5*N); label("\cdots",(5,0),N); label("X_{n-1}",(6.5,0),0.5*N); label("X_n",(7.5,0),0.5*N); label("a_0",(1,0),0.5*S); label("a_1",(2,0),0.5*S); label("a_2",(3,0),0.5*S); label("a_3",(4,0),0.5*S); label("a_{n-2}",(6,0),0.5*S); label("a_{n-1}",(7,0),0.5*S); label("a_n",(8,0),0.5*S); [/asy]$

For all $k \in \mathbb{N}$, the area of $X_k$ is $(a_k - a_{k-1})\left(a_k^{-\dfrac{3}{2}}\right)$. Therefore, $b_n = \sum_{k=1}^{n} [X_k]$

For all sequences $\{ a_k \}$ and all $k \in \mathbb{N}$, $X_k$ lies above the $x$-axis, below the curve $f(x) = x^{-\dfrac{3}{2}}$, and in between the lines $x = 1$ and $x = a_n$, Also, all such rectangles are disjoint.

Thus, $b_n = \sum_{k=1}^{n} [X_k] < \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx = \left[-\dfrac{2}{\sqrt{x}}\right]_{1}^{a_n} = 2 - \dfrac{2}{\sqrt{a_n}} < 2$ as desired.

By choosing $a_k = 1 + k (\Delta x)$, where $\Delta x > 0$, $b_n$ is a Riemann sum for $\int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$. Thus, $\lim_{\Delta x \to 0^+} b_n = \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$.

Therefore, $\lim_{n \to \infty} \left[ \lim_{\Delta x \to 0^+} b_n \right] = \lim_{n \to \infty} \int_{1}^{a_n} x^{-\dfrac{3}{2}} \,dx$ $= \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{a_n}} = \lim_{n \to \infty} 2 - \dfrac{2}{\sqrt{1+n(\Delta x)}} = 2$.

So for any $c \in [0,2)$, we can always select a small enough $\Delta x > 0$ to form a sequence $\{ a_n \}$satisfying the above properties such that $b_n > c$ for large enough $n$ as desired.