Best guess of the asymptotic value of a finite sequence of terms 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 $a_\infty$.
 A: 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}
$
