Finding a sequence $(s_n)$ that does not converge, but $\frac{1}{n}(s_1 + \ldots + s_n)$ does converge. I have the following problem that I am struggling to develop an intuition about.
Let ($s_n$) be a sequence of nonnegative numbers, and for each $n$ define $ \sigma_n = \frac{1}{n}(s_1+s_2+...+s_n)$
I am pretty comfortable with the proof for why lim inf$s_n \leq$ lim inf$\sigma_n$ although I am having trouble thinking of an example where
lim$\sigma_n$ exists but lim$s_n$ does not exist.
 A: Following the suggestion in the comments by @Brian Tung -- a similar suggestion by @Mark Viola would also work -- let $s_n=0$ if $n$ is even and $1$ if $n$ is odd.  Then the sequence ($s_n$) does not converge*, but $\sigma_n=1/2$ if $n$ is even and $1/2-1/2n$ if $n$ is odd, so the sequence $(\sigma_n)$ converges to $1/2$**.
*if there were a limit $L$ then for $\epsilon = 1/2$ there would be $N$ such that $\forall n \ge N,$ $|s_n-L| < 1/2$, but then for $n \ge N$, $|s_n-s_{n+1}|\le |s_n-L|+|s_{n+1}-L| < 1/2 + 1/2,$ contradicting the fact that $|s_{n+1}-s_n|=1$
**for any $\epsilon > 0$, choose $N > 1/2 \epsilon$.  Then $\forall n \ge N,$ $|\sigma_n - 1/2|$, which is $0$ or $1/2n,$ is $\lt \epsilon$.
A: Well, I noted the second expression
$$\sigma_n := \frac{1}{n} (s_1 + \cdots + s_n)$$
as basically being the same expression as used in Cesaro summation, but also more generally is simply the average of the first $n$ terms - and perhaps the averaging interpretation may be more elementary: intuitively, if we imagine $n$ as a growing time index, then what we're looking at is the long-term average of the sequence $(s_n)$, and thus what this question is asking for is really a sequence which does not converge, but nonetheless has a steady, long-term average.
One may be hard-pressed to find such a sequence if one interprets "does not converge" as meaning "grows toward infinity", but one should also note that a sequence can "not converge" if instead of growing, it instead fluctuates in some fashion, with the fluctuation never dying out. One might be led, in real-life terms to thinking about, say, fluctuation in the price of a good on the market, or perhaps more reliably closer to what we're dealing with here, noise at a fixed maximum volume on an electronic signal such as that from your computing device's mike port or speaker with pure background noise.
If we want to construct such a sequence mathematically, one simple way to do so is to exploit the frequency interpretation of probability: this says the probability is, by definition, a kind of long-term average of a sequence. If one considers a sequence of terms that are either $0$ or $1$, chosen perfectly randomly with a $\frac{1}{2}$ probability, i.e. by flipping a truly ideal coin, then this definition implies such a sequence, $s_n$, will have LTA, i.e. the limit of its $\sigma_n$, as being $\frac{1}{2}$, at least "with probability 1" (which basically means that while it's not impossible something else will happen, you'd have to, in a sense, be "infinitely lucky" for something else to happen). An initial part of such a $(s_n)$ might look like
$$s_n: 1, 0, 1, 0, 1, 1, 1, 1, 0, 0, \cdots$$
However, we don't actually need a truly random sequence - this is just for intuition. As long as the sequence contains an even proportion of $1$ and $0$, then its long-term average will be $\frac{1}{2}$, which can be seen by seeing that if (as close to as possible in the case $n$ is odd), we take half the $s_1, \cdots, s_n$ appearing in the definition of $\sigma_n$ to be $1$ and the other half to be $0$ - no matter which ones since addition is commutative - then the $\sigma_n$ should be close to $\frac{1}{2}$.
Thus we can, even more simply, take
$$s_n := \frac{1}{2} \left[(-1)^n + 1\right]$$
.
