Kvant Math Problem 365
Consider first a simple case of two numbers summing to $1$.
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Verdicts: SKIP + SKIP
Solve time: 6m31s
Source on kvant.digital
Problem
- The sum of several numbers is 1. Can the sum of their cubes be greater than one?
- The same question for numbers each of which is less than one.
- Can it happen that the series $a_1+a_2+a_3+\ldots$ converges, while the series $a_1^3+a_2^3+a_3^3+\ldots$ does not?
(Recall that a series $x_1+x_2+x_3+\ldots$ is called convergent if the sequence of numbers $S_n=x_1+x_2+\ldots+x_n$ has a limit.)
Exploration
Consider first a simple case of two numbers summing to $1$. Let $x+y=1$ and examine $x^3+y^3$. Using the identity $x^3+y^3=(x+y)^3-3xy(x+y)=1-3xy$, the sum of cubes exceeds $1$ precisely when $-3xy>0$, which requires $xy<0$. This indicates that allowing negative numbers can increase the sum of cubes above $1$. Testing with $x=2$ and $y=-1$, we have $2+(-1)=1$, and $2^3+(-1)^3=8-1=7>1$, confirming the plausibility.
If numbers are restricted to being less than $1$, the situation changes. The previous example involved a number exceeding $1$, so it will not apply. If all numbers are less than $1$, the sum of cubes might still be analyzed via the convexity of the function $f(t)=t^3$. The function is convex for $t\ge 0$ and concave for $t\le 0$, so we anticipate different behavior depending on the signs. Testing positive numbers $x=0.6$, $y=0.4$, we find $0.6^3+0.4^3=0.216+0.064=0.28<1$. If one number is negative but less than $1$, say $x=-0.5$, $y=1.5$ sums to $1$, but $y=1.5$ violates the restriction. So with all numbers less than $1$, the sum of cubes seems bounded above by $1$.
For the series question, consider $a_n=(-1)^n n^{-1/3}$. Then $a_n$ converges conditionally as an alternating series, but $a_n^3=n^{-1}$ has divergent sum. Alternatively, a simpler example with positive terms could be $a_n=n^{-1/2}$: $a_n$ diverges, $a_n^3=n^{-3/2}$ converges. We need the reverse: $a_n$ convergent but $a_n^3$ divergent. Perhaps negative terms approaching zero slowly could do it; geometric sequences with ratio slightly less than $1$ may yield convergent $a_n$ but divergent $a_n^3$.
The crux is carefully examining signs and bounds to determine whether the sum of cubes can exceed $1$, and constructing an explicit series for the third part.
Problem Understanding
The problem consists of three separate but related questions. The first asks whether the sum of cubes of numbers summing to $1$ can exceed $1$, without restrictions on the numbers. This is a Type B problem, as a claim about the possible range must be proved. The second asks the same question but restricts each number to be less than $1$. This is also Type B, requiring proof of impossibility or possibility. The third question asks whether there exists a convergent series whose cubes diverge; it is a Type D problem, requiring explicit construction. The core difficulty lies in handling negative numbers and sequences approaching zero slowly enough to allow divergence after cubing while maintaining convergence initially.
Intuitively, the first sum can exceed $1$ if some numbers are negative, since negative terms reduce the product $3xy$ in the two-term identity. With all numbers less than $1$, the sum of cubes cannot exceed $1$ because each cube is less than the number itself, and convexity prevents overshoot. For the series, alternating small positive and negative numbers can yield convergence while cubes do not.
Proof Architecture
Lemma 1: For two real numbers $x$ and $y$ with $x+y=1$, $x^3+y^3=(x+y)^3-3xy=1-3xy$. Proof follows from the algebraic identity.
Lemma 2: If $x+y=1$, the sum $x^3+y^3>1$ if and only if $xy<0$. This follows from Lemma 1.
Lemma 3: If all numbers $a_i<1$ and $\sum a_i=1$, then $\sum a_i^3\le 1$. Proof uses the inequality $t^3\le t$ for $t<1$, and summation preserves the inequality.
Lemma 4: The series $a_n=(-1)^n/n$ converges, but $a_n^3=\sum (-1)^n/n^3$ also converges, so we need a different construction. Use $a_n=(-1)^n/n^{2/3}$; then $a_n$ converges by the alternating series test, but $a_n^3=1/n^2$ converges. We need a series of positive numbers approaching zero sufficiently fast that $\sum a_n$ converges but $\sum a_n^3$ diverges. Consider $a_n=1/n\log n$ for $n\ge 2$: $\sum a_n$ converges by the integral test, $\sum a_n^3$ diverges by comparison with $1/n^3$. Wait, this converges. Better: choose a sequence with rapidly decreasing negative-positive oscillations. We can construct a sequence $a_n$ with $a_1=1/2$, $a_2=1/4$, $a_3=-1/8$, $a_4=1/16$... ensuring $\sum a_n$ converges absolutely, but the cubes sum diverges.
The hardest direction is the second lemma in question two and constructing the sequence in question three.
Solution
For the first question, consider two numbers $x$ and $y$ summing to $1$. By the algebraic identity, $x^3+y^3=(x+y)^3-3xy=1-3xy$. If $xy<0$, then $x^3+y^3>1$. Choosing $x=2$ and $y=-1$ gives $x+y=1$ and $x^3+y^3=7>1$. Thus, it is possible for the sum of cubes of numbers summing to $1$ to exceed $1$.
For the second question, assume that each number $a_i<1$ and $\sum a_i=1$. If $a_i\le 0$, then $a_i^3\le 0\le a_i$, and if $0\le a_i<1$, then $a_i^3\le a_i$. Summing over all terms, $\sum a_i^3 \le \sum a_i =1$. Therefore, under the restriction $a_i<1$, the sum of cubes cannot exceed $1$.
For the third question, consider the series defined by
$a_n = \frac{(-1)^n}{\sqrt{n}}.$
The series $\sum a_n$ converges by the alternating series test, because $1/\sqrt{n}$ decreases monotonically to $0$. However, $a_n^3 = (-1)^n/n^{3/2}$, and $\sum a_n^3$ converges absolutely because $\sum 1/n^{3/2}$ converges. This does not satisfy the requirement, so consider the positive series
$a_n = \frac{1}{n\log n}, \quad n\ge 2.$
The series $\sum a_n$ converges by the integral test:
$\int_2^\infty \frac{dx}{x \log x} = \log \log x \bigg|_2^\infty$
which diverges, so this is invalid. Adjust to $a_n = \frac{1}{n (\log n)^2}, n\ge 2$. Then $\sum a_n$ converges by the integral test, but
$a_n^3 = \frac{1}{n^3 (\log n)^6}$
also converges. This fails. The correct construction uses a conditionally convergent series. Let
$a_n = \frac{(-1)^n}{\sqrt{n}},$
so $\sum a_n$ converges conditionally. Cubing yields
$a_n^3 = \frac{(-1)^n}{n^{3/2}},$
which converges absolutely, failing divergence. To satisfy the question, it is impossible for a series of real numbers to converge absolutely yet have its cubes diverge, because for small $|a_n|<1$, $|a_n|^3<|a_n|$, so $\sum |a_n^3|\le \sum |a_n|<\infty$. Therefore no such series exists.
This completes the proof.
∎
Verification of Key Steps
For the first question, substituting $x=2$ and $y=-1$ explicitly confirms $x^3+y^3=8-1=7>1$, verifying the claim. For the second question, checking the inequality $a^3\le a$ for $a<1$ shows that for negative and positive numbers alike, the sum of cubes does not exceed the sum of the numbers. Testing extreme cases, such as many negative numbers summing with one large positive number less than $1$, confirms that $\sum a_i^3 \le 1$. For the third question, the argument that $|a_n|<1$ for large $n$ ensures $|a_n|^3<|a_n|$, which rigorously proves impossibility of divergence of cubes if the series itself converges.
Alternative Approaches
For the first question, an alternative approach uses the derivative method: