Consider an infinite sequence: $$S_\infty := (a_1,a_2,\cdots).$$ Further, suppose that the $a_i$ are all finite and the limit $a_\infty := \lim_{i \to \infty} a_i$ is well-defined and $a_\infty$ is finite. Besides the information above, the $a_i$ are not a priori known.
Suppose now you are being given only the first $n$ numbers of this sequence: $$S_n = (a_1,a_2,\cdots,a_n)$$
is there a mathematical way to have a "best guess" as to what value $a_\infty$ is? For example, if I were to give you the following set of points, plotted in Mathematica:
I would assume most people would (correctly) guess $a_\infty=1$ (the plot is obtained by plotting $1+\frac{\sin x}{x^2}$). Is there a way to make this precise?
In response to @Schach21 comment, I thought about how to make mathematically precise the idea that the sequence does not do anything "crazy". Specifically, the $a_i$ in a way hold information as to what $a_\infty$ can be. The best I could come up with is the following:
In addition, you are told that there exists a set of weights $\{w_i\}_{i=1}^n$, where $w_i \neq 0$ and $\sum_{i=1}^n w_i = 1$, such that the following holds: $$\sum_{i=1}^n w_i a_i = a_\infty.$$ In other words, a well chosen weighted average of the terms in $S_n$ will coincide with $a_\infty$.
Another possibility is exponential weighting.
If $0 < c \lt 1$ then
$\begin{array}\\ A_n &=\dfrac{\sum_{k=1}^n a_kc^{n-k}}{\sum_{k=1}^n c^{n-k}}\\ &=\dfrac{\sum_{k=1}^n a_kc^{n-k}}{\sum_{k=0}^{n-1} c^{k}}\\ &=\dfrac{(1-c)\sum_{k=1}^n a_kc^{n-k}}{1-c^n}\\ \end{array} $
A nice property of this is that it is computationally efficient: $A_{n+1}$ can be gotten from $A_n$ simply.
Let $B_n =\sum_{k=1}^n a_kc^{-k} $ so $A_n =\dfrac{(1-c)c^nB_n}{1-c^n} =\dfrac{(1-c)B_n}{c^{-n}-1} $ or $B_n =A_n\dfrac{c^{-n}-1}{1-c} $. Then $B_{n+1} =B_n+a_{n+1}c^{-n-1} $ so
$A_{n+1}\dfrac{c^{-n-1}-1}{1-c} =A_n\dfrac{c^{-n}-1}{1-c}+a_{n+1}c^{-n-1} $ or
$\begin{array}\\ A_{n+1} &=A_n\dfrac{c^{-n}-1}{1-c}\dfrac{1-c}{c^{-n-1}-1}+a_{n+1}c^{-n-1}\dfrac{1-c}{c^{-n-1}-1}\\ &=A_n\dfrac{c^{-n}-1}{c^{-n-1}-1}+a_{n+1}c^{-n-1}\dfrac{1-c}{c^{-n-1}-1}\\ &=A_n\dfrac{c-c^{n+1}}{1-c^{n+1}}+a_{n+1}\dfrac{1-c}{1-c^{n+1}}\\ &=(1-b_{n+1})A_n+b_{n+1}a_{n+1}\\ \end{array} $
where $b_{n+1} =\dfrac{1-c}{1-c^{n+1}} $.
We can get $b_{n+1}$ in terms of $b_n$, but I don't know how useful this is, since just getting $c^{n+1} =c\cdot c^n$ seems good enough. Anyway, here it is.
Since $1-c^n =\dfrac{1-c}{b_n} $, $c^n =1-\dfrac{1-c}{b_n} $ so $c^{n+1} =c-\dfrac{c(1-c)}{b_n} $ so
$\begin{array}\\ b_{n+1} &=\dfrac{1-c}{1-c^{n+1}}\\ &=\dfrac{1-c}{1-c-\dfrac{c(1-c)}{b_n}}\\ &=\dfrac{1-c}{1-\dfrac{c}{b_n}}\\ &=\dfrac{b_n(1-c)}{b_n-c}\\ \end{array} $