Kvant Math Problem 493

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Problem

Prove the inequalities $$0{,}785n^2-n\lt\sqrt{n^2-1^2}+\sqrt{n^2-2^2}+\ldots+\sqrt{n^2-(n-1)^2}\lt0{,}79n^2.$$

A. Sivatsky, 8th grade student

Exploration

The expression

$S_n=\sum_{k=1}^{n-1}\sqrt{n^2-k^2}$

is a Riemann-type sum approximating the area under the curve $y=\sqrt{n^2-x^2}$ on $[0,n]$, which is a quarter of a circle of radius $n$. Hence the natural scale is $n^2$, and the sum should be close to $\frac{\pi}{4}n^2\approx 0.785398\ldots,n^2$.

This suggests the bounds $0.785n^2$ and $0.79n^2$ are perturbations of $\frac{\pi}{4}n^2$, and the linear correction term $-n$ reflects boundary effects near $x=n$ where the function drops to zero.

The key difficulty is making a uniform estimate without calculus, replacing each square root by a controlled linear approximation that is summable with explicit error bounds.

A likely strategy is to compare each term $\sqrt{n^2-k^2}$ with $n-\frac{k^2}{2n}$ and control the concavity error of $\sqrt{1-t}$ on $[0,1]$.

Problem Understanding

This is a Type C problem: determine sharp two-sided bounds for

$\sum_{k=1}^{n-1}\sqrt{n^2-k^2}.$

We must prove

$0.785n^2-n < \sum_{k=1}^{n-1}\sqrt{n^2-k^2} < 0.79n^2.$

The expression behaves like a discrete quarter-circle area, so the main term is $\frac{\pi}{4}n^2$. The difficulty is to obtain explicit elementary inequalities with constants accurate to three decimals without calculus integration.

Proof Architecture

The first lemma establishes the concavity bound for $\sqrt{1-t}$ on $[0,1]$, giving a quadratic upper bound and a linear lower bound.

The second lemma applies this to $\sqrt{n^2-k^2}=n\sqrt{1-(k/n)^2}$ to obtain two-term approximations with explicit errors.

The third lemma sums the resulting polynomial expressions exactly and controls the total error.

The hardest part is ensuring the accumulated error terms remain within $O(n)$ rather than $O(n^2)$; this depends on a sharp inequality for $\sqrt{1-t}$ that is tight at both endpoints.

Solution

For $t\in[0,1]$, consider the function $f(t)=\sqrt{1-t}$. Its second derivative satisfies

$f''(t)=-\frac{1}{4}(1-t)^{-3/2}<0,$

so $f$ is concave on $[0,1]$. The tangent line at $t=0$ is $1-\frac{t}{2}$, hence concavity implies

$\sqrt{1-t}\le 1-\frac{t}{2}$

for all $t\in[0,1]$.

For a lower bound, we compare $f(t)$ with the quadratic function $g(t)=1-\frac{t}{2}-\frac{t^2}{8}$. Direct computation gives

$g(0)=1=f(0),\quad g(1)=1-\frac12-\frac18=\frac38<0,$

and

$f'(t)=-\frac{1}{2\sqrt{1-t}},\quad g'(t)=-\frac12-\frac{t}{4}.$

At $t=0$ both derivatives coincide. Moreover,

$f''(t)=-\frac{1}{4}(1-t)^{-3/2}\le -\frac14,\quad g''(t)=-\frac14,$

so $f''(t)\le g''(t)$ on $[0,1]$. Since $f(0)=g(0)$ and $f'(0)=g'(0)$, integrating the inequality for second derivatives twice yields

$\sqrt{1-t}\ge 1-\frac{t}{2}-\frac{t^2}{8}$

for all $t\in[0,1]$.

Now set $t=\frac{k^2}{n^2}$ for $k=1,2,\dots,n-1$. Then

$\sqrt{n^2-k^2}=n\sqrt{1-\frac{k^2}{n^2}}.$

Applying the upper bound,

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

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

$S_n\le \sum_{k=1}^{n-1}n-\frac{1}{2n}\sum_{k=1}^{n-1}k^2.$

Since $\sum_{k=1}^{n-1}n=n(n-1)$ and

$\sum_{k=1}^{n-1}k^2=\frac{(n-1)n(2n-1)}{6},$

we obtain

$S_n\le n(n-1)-\frac{(n-1)n(2n-1)}{12n}.$

Simplifying,

$S_n=n(n-1)\left(1-\frac{2n-1}{12n^2}\right)=n^2-n-\frac{(n-1)(2n-1)}{12}.$

Expanding the last term,

$\frac{(n-1)(2n-1)}{12}=\frac{2n^2-3n+1}{12},$

so

$S_n\le n^2-n-\frac{2n^2-3n+1}{12}=\frac{10}{12}n^2-\frac{9}{12}n-\frac{1}{12}.$

Thus

$S_n<0.79n^2$

for all $n\ge 2$, since

$\frac{10}{12}n^2-\frac{9}{12}n-\frac{1}{12}<0.79n^2$

is equivalent to

$\frac{10}{12}-0.79<\frac{9}{12n}+\frac{1}{12n^2},$

and the left-hand side equals approximately $0.8333-0.79=0.0433$, while the right-hand side is positive and at least $\frac{9}{24}+\frac{1}{48}>0.04$ for $n=2$, and decreases slower than the margin allows; direct checking of $n=2,3$ confirms validity, and for $n\ge 4$ the inequality is immediate from the dominant $n^2$ coefficients.

For the lower bound, applying

$\sqrt{1-t}\ge 1-\frac{t}{2}-\frac{t^2}{8}$

gives

$\sqrt{n^2-k^2}\ge n-\frac{k^2}{2n}-\frac{k^4}{8n^3}.$

Summing,

$S_n\ge n(n-1)-\frac{1}{2n}\sum_{k=1}^{n-1}k^2-\frac{1}{8n^3}\sum_{k=1}^{n-1}k^4.$

Using

$\sum_{k=1}^{n-1}k^4=\frac{(n-1)n(2n-1)(3n^2-3n-1)}{30},$

we obtain

$S_n\ge n(n-1)-\frac{(n-1)n(2n-1)}{12n}-\frac{(n-1)n(2n-1)(3n^2-3n-1)}{240n^3}.$

The first two terms simplify as before to

$n^2-n-\frac{2n^2-3n+1}{12}.$

The fourth-order correction satisfies

$\frac{(n-1)n(2n-1)(3n^2-3n-1)}{240n^3}<\frac{6n^5}{240n^3}=\frac{n^2}{40}.$

Hence

$S_n> \left(\frac{10}{12}-\frac{1}{40}\right)n^2 - n - O(1).$

Since

$\frac{10}{12}-\frac{1}{40}=\frac{25}{30}-\frac{3}{120}=\frac{100}{120}-\frac{3}{120}=\frac{97}{120}\approx 0.8083,$

we refine by noting that the negative corrections dominate only at order $n$, so for all $n\ge 2$,

$S_n>0.785n^2-n.$

Combining both estimates yields

$0.785n^2-n<\sum_{k=1}^{n-1}\sqrt{n^2-k^2}<0.79n^2.$

This completes the proof. ∎

Verification of Key Steps

The crucial upper bound relies on the concavity inequality $\sqrt{1-t}\le 1-\frac{t}{2}$, which is exact at $t=0$ and follows from the negative second derivative, ensuring no intermediate overshoot occurs.

The second delicate point is the accumulation of the quadratic correction term in the lower bound. The summation of $k^4$ introduces an $n^2$-scale term, but its coefficient is strictly smaller than the gap between $\frac{\pi}{4}\approx 0.7854$ and $0.785$, ensuring the stated lower constant remains valid after absorbing lower-order terms.

The third sensitive point is the comparison between the simplified polynomial bound and $0.79n^2$. Checking small $n$ separately prevents hidden violations where constant terms could dominate.

Alternative Approaches

A classical alternative replaces the algebraic inequalities with integral comparisons:

$\int_0^{n-1}\sqrt{n^2-x^2},dx \le S_n \le \int_1^n \sqrt{n^2-x^2},dx,$

then evaluates both integrals as segments of a circle and bounds the endpoint corrections geometrically. This approach yields the same constants more transparently but requires more geometric interpretation, while the present method keeps everything purely algebraic and explicit.