MathLabs

Problem 1

Let a0<a1<a2<⋯a_0<a_1<a_2<\cdots be an infinite sequence of positive integers. Prove that there exists a unique integer n≥1n\ge1 such that an<a0+a1+⋯+ann≤an+1.a_n<\frac{a_0+a_1+\cdots+a_n}{n}\le a_{n+1}.
Step 2 of 5: The second inequality is dn+1≤0d_{n+1}\le0
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nan+1−(a0+⋯+an)=−dn+1n a_{n+1} - (a_0+\cdots+a_n) = -d_{n+1}
Detailed analysis

Since (n+1)an+1−(a0+⋯+an+an+1)=−dn+1(n+1)a_{n+1}-(a_0+\cdots+a_n+a_{n+1})=-d_{n+1}, we have nan+1−(a0+⋯+an)=−dn+1na_{n+1}-(a_0+\cdots+a_n)=-d_{n+1}. Hence the inequality a0+⋯+ann≤an+1\frac{a_0+\cdots+a_n}{n}\le a_{n+1} is equivalent to a0+⋯+an≤nan+1a_0+\cdots+a_n\le na_{n+1}, i.e. −dn+1≥0-d_{n+1}\ge0, i.e. dn+1≤0d_{n+1}\le0. So the problem reduces to finding a unique n≥1n\ge1 with dn>0≥dn+1d_n>0\ge d_{n+1}.