Let \( a_0 < a_1 < a_2 \ldots \) be an infinite sequence of positive integers. Prove that there exists a unique integer \( n \geq 1 \) such that

\[ a_n < \frac{a_0 + a_1 + \ldots + a_n } { n} \leq a_{n+1} .\]

Let \( a_0 < a_1 < a_2 \ldots \) be an infinite sequence of positive integers. Prove that there exists a unique integer \( n \geq 1 \) such that

\[ a_n < \frac{a_0 + a_1 + \ldots + a_n } { n} \leq a_{n+1} .\]

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TopNewestLet's call \( S_k = a_0 + a_1 + ... + a_k \) and suppose \[ \frac { { S }_{ k } }{ k } >{ a }_{ k+1 } \; \forall m \leq k \] Then we have \[ k\cdot { S }_{ k }+{ S }_{ k }>k\cdot { a }_{ k+1 }+k\cdot{ S }_{ k }\quad \Leftrightarrow \quad \frac { { S }_{ k } }{ k } >\frac { { S }_{ k+1 } }{ k+1 }\\ \quad \Leftrightarrow \quad \frac { { S }_{ k+1 } }{ k+1 } <\frac { { S }_{ k } }{ k } <\frac { { S }_{ k-1 } }{ k-1 } <...<{ S }_{ 1 }={ a }_{ 0 }+{ a }_{ 1 } \] Hence, we prove by induction \[ 2(a_{ 0 }+{ a }_{ 1 })>{ a }_{ 0 }+{ a }_{ 1 }+{ a }_{ 2 }\quad \Leftrightarrow \quad { a }_{ 0 }+{ a }_{ 1 }>{ a }_{ 2 }\\ k{ a }_{ k+1 }<{ S }_{ k }<k(a_{ 0 }+{ a }_{ 1 })\quad \Leftrightarrow \quad { a }_{ k+1 }<(a_{ 0 }+{ a }_{ 1 }) \] Because \( a_{ 0 }+{ a }_{ 1 } \) is a sum of integer number, there must be a unique number in the sequence for which \( a_{q+1} \geq a_{ 0 }+{ a }_{ 1 } \) and \( a_{q} < a_{ 0 }+{ a }_{ 1 } \). We now show that \( q \) is the required number. We have alredy proved that \[ \frac { { S }_{ q } }{q } \leq { a }_{ q+1 } \] by contraddiction! Thus we have only to show that \[{ a }_{ q }<\frac { S_{ q } }{ q } \quad \Leftrightarrow \quad q{ \cdot a }_{ q }-a_{ q }<S_{ q }-a_{ q }\quad \Leftrightarrow \quad { a }_{ q }<\frac { S_{ q-1 } }{ q-1 } \] And because \[ S_{ q-1 }(q-1)+{ a }_{ q }(q-1)<S_{ q-1 }+S_{ q-1 }(q-1)\quad \Leftrightarrow \quad \frac { S_{ q } }{ q } <\frac { S_{ q-1 } }{ q-1 }\\ \quad \Leftrightarrow \quad \frac { S_{ q } }{ q } <\frac { S_{ q-1 } }{ q-1 } <\frac { S_{ q-2 } }{ q-2 } <...<S_{ 1 }={ a }_{ 0 }+{ a }_{ 1 } \] Excactly as done beofre we must have \( a_q < a_0 + a_1 \). That is true. And we are done. – Andrea Gallese · 3 years ago

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– Scrub Lord · 1 month, 1 week ago

Beautiful solutionLog in to reply

Question 1 was way easier than I expected it to be. Bummed that there were no proper number theory problems. – Mursalin Habib · 3 years ago

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– Calvin Lin Staff · 3 years ago

In recent years, they have made question 1 (and sometimes 4) pretty easy and direct to approach, so that all of the participants (who come from a wide range of ability level) have something to do, and have a chance of geting a HM.Log in to reply

This was indeed very nice, but also pretty easy, even for an IMO P1. I have posted my solution in AoPS.

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=3542324&#p3542324 – Sreejato Bhattacharya · 3 years ago

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Here's a conjecture:

The last possible \(n\) that satisfies \(a_n < \dfrac{a_0+a_1+\cdots +a_n}{n}\) also satisfies \(\dfrac{a_0+a_1+\cdots +a_n}{n}\le a_{n+1}\). This is just a guess though.

These later posts will be what I have came up with. – Daniel Liu · 3 years ago

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So let \(k\) be the maximum such that it still satisfies the condition. Thus, \(a_{k+1} \ge x+y\). However, \(\dfrac{a_0+a_1+\cdots +a_k}{k} < \dfrac{x+y+(x+y)+\cdots +(x+y)}{k}=x+y\). Thus, \(\dfrac{a_0+a_1+\cdots +a_k}{k} < a_{k+1}\).

This seems really sketchy though. Something seems wrong... one of my inequalities was strict when forming the right inequality, but the inequality in the problem is not strict...

EDIT: found out why mine was strict and the problem's wasn't. As long as the maximum \(k\) satisfies \(k\ge 2\), then the upper bound inequality is strict, just like my result. However, the special case of the maximum \(k\) is \(k=1\) yields a non-strict inequality.

TO DO: prove uniqueness of satisfying both inequalities. – Daniel Liu · 3 years ago

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– Daniel Liu · 3 years ago

I realize that I have to tighten the bound for each \(a_k < x+y\) to something more strict, or something like that in order to prove the uniqueness of that one \(k\) that satisfies both inequalities. I feel like I'm almost there, but just can't get over the last little bump. I need to somehow use the \(a_k < a_{k+1}\) to my advantage...Log in to reply