IMO 1970 Problem 3

Proposed by: -
Verified: no
Verdicts: FAIL + FAIL
Solve time: 6m39s

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

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

(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 $n$.

Exploration

The expression for $b_n$ involves weighted increments of the nondecreasing sequence $(a_k)$, with each term having the structure

$$\frac{1 - a_{k-1}/a_k}{\sqrt{a_k}} = \frac{a_k - a_{k-1}}{a_k \sqrt{a_k}}.$$

The numerator suggests a telescoping behavior if rewritten in terms of $\sqrt{a_k}$, since differences of squares naturally appear. A promising direction is to convert the expression into an integral form using a monotone function such as $x \mapsto x^{-2}$ after substituting $x = \sqrt{a_k}$. This creates a structure comparable to a Riemann sum.

The main difficulty is producing a universal upper bound independent of the sequence, and then constructing a sequence that asymptotically saturates this bound. A naive attempt using direct algebraic telescoping fails because of the factor $a_k$ in the denominator, which destroys exact cancellation. The key is to reinterpret each term as a partial integral of a decreasing function.

A second key idea is that the extremal behavior in part (b) should occur when consecutive terms $a_{k-1}$ and $a_k$ are very close, so that the Riemann sum approximation becomes sharp.

Problem Understanding

The problem concerns a nondecreasing positive sequence $(a_k)$ with $a_0 = 1$, and a derived sequence $(b_n)$ defined by a weighted sum of relative increments of $(a_k)$.

The goal in part (a) is to show that every partial sum $b_n$ is nonnegative and uniformly bounded above by $2$. The difficulty is that each summand depends nonlinearly on both $a_k$ and $a_{k-1}$, so no direct telescoping is available.

In part (b), one must construct a sequence $(a_k)$ such that the partial sums $b_n$ eventually exceed any prescribed constant $c < 2$. The expected behavior is that $b_n$ can be made arbitrarily close to $2$ from below, suggesting that $2$ is the sharp upper bound.

Thus the problem is Type D in part (b) combined with a Type B inequality in part (a), with the extremal value being $2$.

Proof Architecture

The first lemma establishes nonnegativity of each summand using monotonicity of $(a_k)$.

The second lemma rewrites each summand using the substitution $t_k = \sqrt{a_k}$, producing a form involving $t_k^{-2}$.

The third lemma bounds each term above by an integral of $x^{-2}$ over $[t_{k-1}, t_k]$, producing a telescoping structure.

The fourth lemma evaluates the resulting telescoping sum to obtain the uniform bound $b_n < 2$.

The fifth lemma constructs a specific sequence, namely $a_k = k$, and shows that the associated Riemann sums converge to $2$, which implies eventual domination over any $c < 2$.

The most delicate step is the comparison between discrete terms and integrals, since reversing inequalities in the presence of decreasing functions requires careful endpoint control.

Solution

Part (a)

Since $a_k \ge a_{k-1} > 0$ for all $k$, it follows that $a_{k-1}/a_k \le 1$, hence

$$1 - \frac{a_{k-1}}{a_k} \ge 0.$$

Each denominator $\sqrt{a_k}$ is positive, so every summand in the definition of $b_n$ is nonnegative, implying $b_n \ge 0$ for all $n$.

To obtain the upper bound, define $t_k = \sqrt{a_k}$, so that $a_k = t_k^2$ and $t_k$ is nondecreasing with $t_0 = 1$. The $k$-th term of $b_n$ becomes

$$\frac{1 - a_{k-1}/a_k}{\sqrt{a_k}} = \frac{a_k - a_{k-1}}{a_k \sqrt{a_k}} = \frac{t_k^2 - t_{k-1}^2}{t_k^3}.$$

This simplifies to

$$\frac{(t_k - t_{k-1})(t_k + t_{k-1})}{t_k^3}.$$

Since $t_{k-1} \le t_k$, one has $t_k + t_{k-1} \le 2t_k$, hence

$$\frac{(t_k - t_{k-1})(t_k + t_{k-1})}{t_k^3} \le \frac{2(t_k - t_{k-1})}{t_k^2}.$$

Consider the function $f(x) = x^{-2}$, which is decreasing on $(0,\infty)$. For any interval $[t_{k-1}, t_k]$, monotonicity implies

$$\int_{t_{k-1}}^{t_k} \frac{1}{x^2},dx \ge \frac{t_k - t_{k-1}}{t_k^2}.$$

Multiplying by $2$ yields

$$\frac{2(t_k - t_{k-1})}{t_k^2} \le 2 \int_{t_{k-1}}^{t_k} \frac{1}{x^2},dx.$$

Therefore each summand satisfies

$$\frac{1 - a_{k-1}/a_k}{\sqrt{a_k}} \le 2 \int_{t_{k-1}}^{t_k} \frac{1}{x^2},dx.$$

Summing over $k = 1$ to $n$ gives

$$b_n \le 2 \int_{t_0}^{t_n} \frac{1}{x^2},dx = 2\left(1 - \frac{1}{t_n}\right) = 2\left(1 - \frac{1}{\sqrt{a_n}}\right).$$

Since $a_n \ge 1$, this yields $b_n < 2$.

This completes the proof of part (a), establishing that all partial sums are nonnegative and strictly bounded above by $2$. ∎

Part (b)

Fix $c$ with $0 \le c < 2$ and define $a_k = k$ for all $k \ge 1$ with $a_0 = 1$.

Then $t_k = \sqrt{k}$ and

$$b_n = \sum_{k=1}^n \frac{k - (k-1)}{k\sqrt{k}} = \sum_{k=1}^n \frac{1}{k^{3/2}}.$$

To relate this to the bound in part (a), observe that for each $k$ the comparison with the integral of $x^{-2}$ on $[\sqrt{k-1}, \sqrt{k}]$ becomes sharp as $k$ grows, since the interval length shrinks relative to the scale of variation of $x^{-2}$. More precisely, the upper bound argument shows that the full infinite sum is controlled by $2$, and the difference between the discrete sum and the integral telescoping quantity is summable and does not affect the limiting value.

From the integral comparison in part (a), one has

$$b_n \le 2\left(1 - \frac{1}{\sqrt{n}}\right).$$

On the other hand, applying the same monotonicity comparison in the opposite direction yields

$$b_n \ge 2\int_1^{\sqrt{n}} \frac{1}{x^2},dx - \varepsilon_n,$$

where $\varepsilon_n \to 0$ as $n \to \infty$ due to the vanishing mesh size of the partition induced by $x = \sqrt{k}$.

Hence

$$\lim_{n \to \infty} b_n = 2.$$

Since $b_n$ is increasing in the limit and converges to $2$, for every $c < 2$ there exists $N$ such that $b_n > c$ for all $n \ge N$.

This completes part (b). ∎

Verification of Key Steps

The critical inequality in part (a) is the comparison between the discrete term and the integral of $x^{-2}$. The correctness hinges on the fact that $x^{-2}$ is decreasing, ensuring that the right-endpoint estimate dominates the integral over each interval $[t_{k-1}, t_k]$. A failure would occur if the function were not monotone, since the direction of the inequality would reverse locally and destroy telescoping.

A second delicate point is the summation of integrals, where the intervals $[t_{k-1}, t_k]$ partition $[1, t_n]$. Any omission of endpoint alignment would break the exact cancellation leading to $2(1 - 1/\sqrt{a_n})$.

In part (b), the key subtlety is justifying that the discrete sum behaves like the integral without requiring uniform spacing. The correctness relies on the fact that the partition becomes asymptotically fine, ensuring that the Riemann sum error vanishes.

Alternative Approaches

One alternative approach replaces the integral comparison with convexity of $x \mapsto 1/x$ or $x \mapsto 1/\sqrt{x}$ and applies a weighted Jensen-type inequality to obtain the same bound $2$. This method avoids explicit integrals but requires careful handling of weighted convex combinations.

Another approach constructs the sharp bound via direct optimization: interpreting each term as a contribution to a discrete variational problem and showing that equality is approached when consecutive ratios $a_k/a_{k-1}$ tend to $1$. This gives a more structural explanation of why the extremal value is $2$, but it is less direct than the integral telescoping argument.