Kvant Math Problem 532
For small values,
Verified: no
Verdicts: SKIP + SKIP
Solve time: 3m44s
Source on kvant.digital
Problem
Let $a_n=\sqrt{n+1}+\sqrt{n\vphantom2}$ and $b_n=\sqrt{4n+2}$. Prove that for every positive integer $n$:
- $[a_n]=[b_n]$, where $[x]$ denotes the integer part of the number $x$;
- $0\lt b_n-a_n\lt\dfrac1{16n\sqrt n}$.
Exploration
For small values,
$$a_1=\sqrt2+1\approx2.414,\qquad b_1=\sqrt6\approx2.449,$$
so $[a_1]=[b_1]=2$ and
$$b_1-a_1\approx0.035.$$
For $n=2$,
$$a_2=\sqrt3+\sqrt2\approx3.146,\qquad b_2=\sqrt{10}\approx3.162,$$
again the integer parts coincide.
To compare $a_n$ and $b_n$, compute their squares:
$$a_n^2=(\sqrt{n+1}+\sqrt n)^2=2n+1+2\sqrt{n(n+1)},$$
$$b_n^2=4n+2.$$
The difference is
$$b_n^2-a_n^2 =2n+1-2\sqrt{n(n+1)} =(\sqrt{n+1}-\sqrt n)^2.$$
Hence $b_n>a_n$. This already suggests that the two numbers are extremely close, because the difference of their squares is of order $1/n$.
To prove equality of integer parts, it is enough to show that no integer lies between $a_n$ and $b_n$. Since $b_n>a_n$, a natural idea is to prove $b_n-a_n<1$. The given stronger estimate should accomplish this.
The crucial point is obtaining the sharp upper bound. Using
$$b_n-a_n=\frac{b_n^2-a_n^2}{b_n+a_n} =\frac{(\sqrt{n+1}-\sqrt n)^2}{b_n+a_n},$$
the numerator can be rewritten as
$$(\sqrt{n+1}-\sqrt n)^2 =\frac1{(\sqrt{n+1}+\sqrt n)^2} =\frac1{a_n^2}.$$
Then
$$b_n-a_n=\frac1{a_n^2(a_n+b_n)}.$$
Since $a_n>2\sqrt n$ and $a_n+b_n>4\sqrt n$, the denominator exceeds
$$(2\sqrt n)^2\cdot4\sqrt n=16n\sqrt n,$$
which gives exactly the desired estimate.
For the equality of integer parts, once
$$0<b_n-a_n<\frac1{16n\sqrt n}\le\frac1{16}<1$$
is known, any integer $m$ satisfying $a_n<m\le b_n$ would force $b_n-a_n\ge1$, impossible.
Problem Understanding
We are given
$$a_n=\sqrt{n+1}+\sqrt n, \qquad b_n=\sqrt{4n+2},$$
and must prove two statements for every positive integer $n$.
First, the integer parts of $a_n$ and $b_n$ are equal.
Second, the difference between the numbers is positive and satisfies
$$0<b_n-a_n<\frac1{16n\sqrt n}.$$
This is a Type B problem, since both statements are prescribed and must be proved.
The core difficulty is obtaining a sufficiently precise expression for $b_n-a_n$. The identity
$$b_n^2-a_n^2=(\sqrt{n+1}-\sqrt n)^2$$
reveals that the difference of the squares is very small, and after rationalization it yields the required estimate.
Proof Architecture
First prove that
$$b_n^2-a_n^2=(\sqrt{n+1}-\sqrt n)^2.$$
This follows from direct expansion of $a_n^2$.
Next prove that $b_n>a_n$.
Since the preceding square difference is positive and both numbers are positive, their order follows.
Then derive the exact identity
$$b_n-a_n=\frac1{a_n^2(a_n+b_n)}.$$
This comes from rationalizing $b_n-a_n$ and using
$$(\sqrt{n+1}-\sqrt n)^2=\frac1{a_n^2}.$$
After that prove
$$a_n>2\sqrt n, \qquad a_n+b_n>4\sqrt n.$$
These inequalities are immediate from the definitions and positivity.
Substituting them into the exact identity yields
$$0<b_n-a_n<\frac1{16n\sqrt n}.$$
Finally show that $[a_n]=[b_n]$.
Since $0<b_n-a_n<1$, no integer can lie strictly between $a_n$ and $b_n$, and therefore the two numbers have the same integer part.
The most delicate lemma is the exact formula
$$b_n-a_n=\frac1{a_n^2(a_n+b_n)},$$
because any algebraic error there would invalidate the sharp estimate.
Solution
We first compare the squares of $a_n$ and $b_n$.
From the definition,
$$a_n^2 =(\sqrt{n+1}+\sqrt n)^2 =2n+1+2\sqrt{n(n+1)},$$
while
$$b_n^2=4n+2.$$
Hence
\begin{align*}
b_n^2-a_n^2
&=4n+2-(2n+1+2\sqrt{n(n+1)})\
&=2n+1-2\sqrt{n(n+1)}\
&=(\sqrt{n+1}-\sqrt n)^2.
\end{align*}
Since $(\sqrt{n+1}-\sqrt n)^2>0$, we obtain
$$b_n^2>a_n^2.$$
Both $a_n$ and $b_n$ are positive, therefore
$$b_n>a_n.$$
This proves the left inequality in part 2.
Now rationalize the difference:
$$b_n-a_n =\frac{b_n^2-a_n^2}{b_n+a_n} =\frac{(\sqrt{n+1}-\sqrt n)^2}{b_n+a_n}.$$
Since
$$\sqrt{n+1}-\sqrt n =\frac1{\sqrt{n+1}+\sqrt n} =\frac1{a_n},$$
we have
$$(\sqrt{n+1}-\sqrt n)^2=\frac1{a_n^2}.$$
Therefore
$$b_n-a_n=\frac1{a_n^2(a_n+b_n)}.$$
Next,
$$a_n=\sqrt{n+1}+\sqrt n>2\sqrt n,$$
so
$$a_n^2>4n.$$
Also,
$$b_n=\sqrt{4n+2}>2\sqrt n,$$
hence
$$a_n+b_n>4\sqrt n.$$
Substituting these inequalities into the exact formula gives
$$b_n-a_n =\frac1{a_n^2(a_n+b_n)} < \frac1{(4n)(4\sqrt n)} = \frac1{16n\sqrt n}.$$
Together with $b_n>a_n$, this yields
$$0<b_n-a_n<\frac1{16n\sqrt n}.$$
Part 2 is proved.
For part 1, observe that
$$\frac1{16n\sqrt n}\le\frac1{16}<1,$$
so part 2 implies
$$0<b_n-a_n<1.$$
Assume that $[a_n]\ne[b_n]$. Since $a_n<b_n$, there exists an integer $m$ such that
$$a_n<m\le b_n.$$
Then
$$b_n-a_n>(m)-(m-1)=1,$$
a contradiction to $b_n-a_n<1$.
Hence no integer lies between $a_n$ and $b_n$. Since $a_n<b_n$, the two numbers must belong to the same unit interval $[k,k+1)$ for some integer $k$, and therefore
$$[a_n]=[b_n].$$
Both required statements have been established. This completes the proof.
∎
Verification of Key Steps
The first delicate step is the computation of $b_n^2-a_n^2$.
Starting from
$$a_n^2=2n+1+2\sqrt{n(n+1)},$$
we obtain
$$b_n^2-a_n^2 =4n+2-2n-1-2\sqrt{n(n+1)} =2n+1-2\sqrt{n(n+1)}.$$
Expanding
$$(\sqrt{n+1}-\sqrt n)^2 =n+1+n-2\sqrt{n(n+1)} =2n+1-2\sqrt{n(n+1)},$$
which matches exactly.
The second delicate step is the identity
$$(\sqrt{n+1}-\sqrt n)^2=\frac1{a_n^2}.$$
Since
$$(\sqrt{n+1}-\sqrt n)(\sqrt{n+1}+\sqrt n)=1,$$
we have
$$\sqrt{n+1}-\sqrt n=\frac1{\sqrt{n+1}+\sqrt n} =\frac1{a_n}.$$
Squaring yields the required formula. Forgetting to square the denominator would produce an incorrect estimate.
The final delicate step is deducing equality of integer parts from the bound. The argument requires both inequalities
$$a_n<b_n \quad\text{and}\quad b_n-a_n<1.$$
If $a_n<b_n$ were not established, the existence of an integer between them would not follow from differing integer parts. The positivity of the difference is therefore an essential input.
Alternative Approaches
After proving $b_n>a_n$, one may compare both numbers directly with consecutive integers. Let
$$k=[b_n].$$
From
$$k^2\le4n+2<(k+1)^2$$
one can derive bounds on $n$ and then show that $a_n$ also lies in $[k,k+1)$. This proves the equality of integer parts without using the estimate from part 2. The calculations are longer and less transparent.
Another route to part 2 is to write
$$b_n-a_n =\sqrt{4n+2}-\sqrt{4n+1-\bigl(2\sqrt{n(n+1)}-2n\bigr)}$$
and apply the mean value theorem to the function $\sqrt{x}$. This yields a bound of the correct order, but the exact identity obtained by rationalization gives the estimate immediately and with much less computation.