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Let $\{a_n\}_{n=1}^{\infty}$ be a sequence such that $a_i\in[0,1]$ for every $i\in \mathbb{N}$, and suppose that $$\lim_{n \to \infty}\frac{1}{n} \sum_{i=1}^n a_i = p.$$

Does $$\frac{1}{n} \sum_{i=1}^n a_i^2$$ necessarily converge when $n \to \infty$?

Clearly, that fact that $$\left|\frac{1}{n} \sum_{i=1}^n a_i\right|\geqslant\left|\frac{1}{n} \sum_{i=1}^n a_i^2\right|$$ is not enough, but together with the fact that the sequence $a_n^2$ is the square of $a_n$, is it enough?

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  • $\begingroup$ What does $\{a_n\}_{n=1}^\infty$ mean? Is this equal to the set $\{a_1, a_2, a_3,\ldots\}$? I have never seen such notation before. $\endgroup$ – Mr Pie Feb 15 '18 at 13:00
  • $\begingroup$ @user477343 It is a rather commong notation for a sequence $a_1,a_2,...$ $\endgroup$ – M. Winter Feb 15 '18 at 13:00
  • $\begingroup$ @M.Winter thank you for telling me. You learn something new everyday... $\endgroup$ – Mr Pie Feb 15 '18 at 13:01
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No. Observe that $$\frac{1+0}2=\frac12=\frac{2/3+1/3}2$$ but $$\frac12=\frac{1^2+0^2}2\ne\frac{(2/3)^2+(1/3)^2}2=\frac5{18}.$$

Take $(a_n)$ to be a sequence constructed like so: A long stretch of $1,0,1,0,\dots,1/0$ followed by a much longer stretch of $2/3,/1/3,\dots,2/3,1/3$ followed by an even longer sequence of alternating $1$s and $0$s followed by a truly awesomely long sequence of alternating $2/3$s and $1/3$s, etc.

Then $\frac1n\sum_1^n a_j\to1/2$ regardless of how long we take the alternating subsequences. But no matter what we've done so far, if we append a long enough sequence of $1,0$ pairs we reach a point where $\frac1n\sum_1^na_j^2$ is close to $1/2$, and then if we append a long enough sequence of $2/3,1/3$ pairs we reach a point where $\frac1n\sum_1^na_j^2$ is close to $5/18$. Then we get close to $1/2$ again and then close to $5/18$ again, etc, so $\frac1n\sum_1^na_j^2$ does not converge.

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