A number sequence $a_1, a_2, a_3….,a_n$ is such that $a_1 = 0; |a_2| = |a_1+1|; |a_3| = |a_2+1|; |a_4| = |a_3+1|; …. ; |a_{n+1}| = |a_n+1|$. Prove that $AM$ of of $a_1, a_2, a_3….,a_n$ cannot be lesser than $\frac{-1}{2}$.
Approach:
$$a_1 = 0$$
$$\Longrightarrow |a_2| = 1; a_2 = \pm1$$
$$a_2 = \pm1$$
$$\Longrightarrow |a_3| = 2,0; a_3 = \pm2,0$$
$$a_3 = \pm2,0$$
$$\Longrightarrow |a_4|=3,1; a_4 = \pm3, \pm1$$
I got the terms as above.
We need to prove that: $$\cfrac{a_1+ a_2+ a_3+….+a_n}{n}\geqslant\cfrac{-1}{2}$$
Rather than beginning with a proof, I decided to go backwards, so:
$${a_1+ a_2+ a_3+….+a_n}\geqslant\cfrac{-n}{2}$$
And to get to it in a fairly easy way I assumed the series to be as: $0,-1,0,-1,0,-1….$ from the terms I found.
When $n$ is odd:
$$\cfrac{0-1+0-1+0-1….0}{n} = \cfrac{\frac{-(n-1)}{2}}{n} = \cfrac{-n-1}{2n} = \cfrac{-1}{2} + \cfrac{1}{2n} > \cfrac{-1}{2}$$
When $n$ is even:
$$\cfrac{0-1+0-1+0-1….-1}{n} = \cfrac{\frac{-(n)}{2}}{n} = \cfrac{-1}{2}$$
This is kind of cheap, I don't know how to generalize this or prove it through induction either.
I'm looking for a proof involving the summation operator or induction. Any help would be appreciated!
Best Answer
We have $$ a_{n+1}^2 = (a_n+1)^2 = a_n^2 + 2a_n + 1 $$ for all $n$. It follows that $$ \sum_{k=1}^n (2a_k+1) = \sum_{k=1}^n (a_{k+1}^2 - a_k^2) \, . $$ The right-hand side is a telescoping sum, and we get $$ n + 2 \sum_{k=1}^n a_k = a_{n+1}^2 \ge 0 \, . $$
Remark: Your observation that $$ a_1 = 0, \, a_2 = \pm 1, \,a_3 = \pm 2, 0, \,a_4 = \pm 3, \pm 1 \, \ldots $$ leads to the conjecture that $|a_n| \le n-1$, and therefore $a_n \ge 1-n$. That is indeed true, and can be proved with induction. But it does not give the desired estimate because not all combinations are possible. As an example, $a_4=-3$ is only possible if $a_2=1$ and $a_3=2$.