5

Let $a_n$ be a sequence of positive numbers such that $a_0=1$ and $\forall n : a_n \ge a_{n+1}$.

Find the infinum of $\sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}}$ over all such sequences.

If $\{a_n\}$ is a geometric seriess $1, q, q^2, ...$, where $0<q<1$, then the sum equals $\frac{1}{q}+\frac{1}{1-q}$, which has minimum at $q=\frac{1}{2}$, so the infinum is no greater than $4$.

I suspect that the answer is indeed $4$, but I have no idea how to prove that.

kvardekkvar
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1 Answers1

2

The infimum is indeed $4$.

We have $$ \begin{align} \sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}}&\geq \sum_{i=0}^n \frac{a_i^2}{a_{i+1}}+\sum_{i=n+1}^{\infty} a_i \ \ \ \ \textrm{( by $a_i\geq a_{i+1}$: replace denominator.)}\\ &>\sum_{i=0}^n \frac{a_i^2}{a_{i+1}} + a_{n+1} \ \ \ \ \textrm{(drop all terms with $i\geq n+2$.)}\\ &\geq \sum_{i=0}^{n-1} \frac{a_i^2}{a_{i+1}}+2a_n \ \ \ \ \textrm{(apply AM-GM to the last two terms)}\\ &\geq \sum_{i=0}^{n-2} \frac{a_i^2}{a_{i+1}}+2\sqrt 2 a_{n-1} \ \ \textrm{(apply AM-GM to the last two)} \end{align} $$ We repeat this process until we obtain $$ \sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}} > \frac1{a_1}+2^{1+\frac12+\cdots + \frac1{2^{n-1}}}a_{n-(n-1)}. $$ Then applying AM-GM to this one, we obtain $$ \sum_{i=0}^{\infty} \frac{a_i^2}{a_{i+1}} > 2^{1+\frac12+\cdots+\frac1{2^n}} $$ This lower bound can be made arbitrary close to $4$. Hence, the claim is proved.

Sungjin Kim
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