A bounded sequence whose sequence of averages does not converge

Question

Can we find a bounded sequence $\{a_n\}$ such that the sequence of its averages, say, sequence $\{b_n\}$, where $$b_n=\frac{1}{n}\sum_{i=1}^n a_i,$$ does not converge?

Answer 1

Sure. Select $a_n\in\{-1,+1\}$ in such a manner that $b_n$ dances (slower and slower) between $-1$ and $+1$.

More explicitly (though less extreme): Let $a_n=(-1)^{\lfloor \log_2 n\rfloor}$. Then at $n=2^m$, we have $\sum_{k=1}^n a_k\approx (-1)^m\cdot \frac n3$ because the ratio of the different values is approximately $2:1$ with changing leadership, hence $b_n\approx (-1)^m\frac13$.

Answer 2

One way is to have a sequence of zeroes and ones. You add $1$'s to your sequence until the average goes above say ${3 \over 4}$, then add $0$'s to your sequence until the average goes below ${1 \over 4}$, then add more $1$'s until it goes above ${3 \over 4}$, and back to zeroes ad infinitum.

A useful fact in showing this works: If $n = n_1 + n_2$, then $${1 \over n} \sum_{k = 1}^n a_k = {n_1 \over n} ({1 \over n_1} \sum_{k = 1}^{n_1} a_k) + {n_2 \over n} ({1 \over n_2} \sum_{k = n_1 + 1}^{n} a_k)$$

So if $A_k$ denotes the average of the first $k$ members of your sequence and $B_{n_1,n}$ denotes the average of the $n_1 + 1$ through $n$th members you have $$A_n = {n_1 \over n} A_{n_1} + {n_2 \over n} B_{n_1,n}$$ $$= {n_1 \over n} A_{n_1} + (1 - {n_1 \over n}) B_{n_1,n}$$ So by increasing $n$ for a given $n_1$, your average $A_n$ becomes a weighted average of $A_{n_1}$ and $B_{n_1,n}$ that emphasizes $B_{n_1,n}$ more and more.

Answer 3

Yes. Think about a series that has a short run of $1$'s, a longer run of $-1$'s, an even longer run of $1$'s, etc. If you make the lengths work correctly, you can get there.