# Showing an infinite sequence is constant under some condition

Let $$a_1,a_2,...$$ be an infinite sequence of positive real numbers such that for each positive integer $$n$$ we have

$$\frac{a_1+a_2+..+a_n}n\ge\sqrt{\frac{a^2_1+a^2_2+...+a^2_{n+1}}{n+1}}.$$

Prove that the sequence $$a_1,a_2,...\$$ is constant.

MY ATTEMPT/THOUGHTS:

My initial plan is to show that the sequence is bounded and then proving it is constant.

For that I considered the following.

Let $$m_n=\min\{a_1,a_2,...,a_n\}$$, $$M_n=\max\{a_1,a_2,...,a_n\}$$, and $$S_n=\frac{a^2_1+a^2_2+...+a^2_n}{n}.$$

Then we have $$m^2_n\le S_{n+1} \le M^2_n.$$

Also from the given inequality, we have, on squaring,

$$\frac{1}{n}S_n+2\frac{a_1a_2+a_1a_3+....+a_{n-1}a_n}{n^2}\ge S^2_{n+1}.$$

I have no idea how to proceed after this or even if I am moving in the right direction!

Do you have any suggestions? Thanks for your time.

• A slightly improved version of this start is as follows. Note that the quadratic mean of real numbers is always at least their arithmetic mean $$\sqrt{\frac{a_1^2+a_2^2+\dots+a_{n+1}^2}{n+1}}\ge\frac{a_1+a_2+\dots+a_{n+1}}{n+1}.$$ Moreover, for nonnegative $a_i$'s, the equality is only possible if all the $a_i$'s are equal. Thus, from the inequality given in the problem, $$\frac{a_1+a_2+\dots+a_{n}}{n}\ge\frac{a_1+a_2+\dots+a_{n+1}}{n+1},$$ i.e. $$a_{n+1}\le\frac{a_1+a_2+\dots+a_{n}}{n}.$$ Commented Jul 22, 2020 at 8:00
• @AlexanderBurstein Thanks sir. Well, based on your hint and my notation, we can say $\{M_n\}$ is non-increasing since otherwise if $a_{n+1} \gt a_i, i=1,2,..,n$ then $na_{n+1} \gt \sum_{i=1}^n a_i$ , a contradiction. This proves the sequence $\{M_n\}$ is bounded above by $a_1$ and hence $a_n$ is bounded for all $n$ . Is this correct ? Then what to do ? Commented Jul 22, 2020 at 11:43
• @AlexanderBurstein Can something be done correspoding to the minimum. ? I mean ,we have already shown $a_n \le a_1 \forall n$ . So we should prove the reverse inequality , shouldn't we ? Commented Jul 22, 2020 at 12:00
• Not sure what to do for the other direction. That's the difficult part. Commented Jul 22, 2020 at 20:03
• – Sil
Commented Sep 19, 2020 at 19:39

We have $$\underbrace{\text{QM}(a_1,\ldots,a_n)}_{Q(n)}\geq \underbrace{\text{AM}(a_1,\ldots,a_n)}_{A(n)} \geq \underbrace{\text{QM}(a_1,\ldots,a_{n+1})}_{Q(n+1)}\geq \underbrace{\text{AM}(a_1,\ldots,a_{n+1})}_{A(n+1)} \tag{0}$$ so both $$A(n)$$ and $$Q(n)$$ are non-increasing and $$a_{n+1}\leq A(n)$$. The central inequality can be written as

$$a_{n+1}^2 \leq (n+1)A(n)^2 - nQ(n)^2 \tag{1}$$ so we must have $$A(n)^2 \geq \frac{n}{n+1} Q(n)^2,\qquad A(n)\geq Q(n)\sqrt{1-\tfrac{1}{n+1}}.$$ We may consider that the average value of $$a_1,\ldots,a_n$$ is $$A(n)$$ and $$V(n)=\frac{1}{n}\sum_{k=1}^{n}(a_k-A(n))^2 = Q(n)^2-A(n)^2\leq \frac{Q(n)^2}{n+1}.$$ $$Q(n)$$ is non-increasing and $$\frac{1}{n+1}$$ is decreasing to zero, so the variance goes to zero as $$n\to +\infty$$.
We may write $$(1)$$ as

$$a_{n+1}^2 \leq A(n)^2 - nV(n) \tag{2}$$

and define a sequence in the following way:

$$a_1=2,\quad a_2=1,\quad a_{n+1}=\sqrt{A(n)^2-nV(n)}$$

$$\{a_n\}_{n\geq 1}=\left\{2,1,\frac{\sqrt{7}}{2},\frac{1}{6} \sqrt{48 \sqrt{7}-71},\frac{1}{12} \sqrt{\frac{15}{2} \sqrt{979+1212 \sqrt{7}}+3 \sqrt{7}-293},\ldots\right\}$$ This seems to work for a few terms, but at some point $$n V(n)=\sum_{k=1}^{n}(a_k-A(n))^2$$ becomes larger than $$A(n)^2$$. Now we have to prove that unless $$\{a_n\}_{n\geq 1}$$ is constant we cannot avoid this phenomenon.

$$\begin{eqnarray*} (n+1)V(n+1)-n V(n) &=& (n+1)Q(n+1)^2-(n+1)A(n+1)^2-n Q(n)^2+n A(n)^2\\&=&(a_{n+1}-A(n+1))^2+n(A(n)-A(n+1))^2\end{eqnarray*}$$ shows that $$n V(n)$$ is weakly increasing.

$$(n+1)V(n+1)=\sum_{k=1}^{n}((k+1)V(k+1)-k V(k))\geq \sum_{k=1}^{n}k(A(k)-A(k+1))^2$$ and $$n\sum_{k=1}^{n}k(A(k)-A(k+1))^2\stackrel{\text{CS}}{\geq}\left(\sum_{k=1}^{n}\sqrt{k}(A(k)-A(k+1))\right)^2$$ can be lower-bounded by using summation by parts:

$$\sum_{k=1}^{n}\sqrt{k}(A(k)-A(k+1)) \geq (A(1)-A(n+1))\sqrt{n}.$$

• Sorry to put a late question. Why does the variance of a sequence converging to $0$ implies the sequence is convergent ? Also defining the sequence in the way you have assures inequality $(2)$ is satisfied but how to deduce or conclude in a backward process that the inequality $(0)$ is satisfied ? I mean the implications are forward directional only, aren't they ? There might be other undiscovered restrictions on the sequence $\{a_n\}$ making it constant. I am not able to convince myself on these things . It will be highly helpful if you clarify these things. Thanks sir. Commented Jul 22, 2020 at 16:44
• Indeed this answer is flawed: at some point $n V(n)> A(n)^2$, unless all the terms are constant. Commented Jul 22, 2020 at 16:51
• Why would we do best with the equality in the recursive definition of the sequence you defined at the end? Thanks. Commented Jul 22, 2020 at 17:14
• Do you think the conclusion that the sequence is constant is specific to AM and QM, or could this be generalized to any $\left(\frac{1}{n}\sum_{k=1}^{n}{a_k^p}\right)^{1/p}\ge\left(\frac{1}{n+1}\sum_{k=1}^{n+1}{a_k^q}\right)^{1/q}$, where $p>q>0$ (or, perhaps, $p>q\ge 1$)? Commented Jul 22, 2020 at 19:59
• Correcting my previous comment: "... $q>p>0$ (or, perhaps, $q>p\ge 1$)?" Commented Jul 22, 2020 at 21:50