0

Let's say we have some sequence of real numbers $\{a_n\}$ which converges to some value $A$.

How does $\frac{\sum{\{a_n\}}}{n}$ behave as $n$ tends to $\infty$? Is it $0$ as the denominator becomes incredibly large or will it go towards $A$?

My thinking is that it can't tend towards $A$ as the earlier terms of the sequence would not be able to "compensate" for the difference between themselves and $A$ (assuming that the sequence is strictly increasing or decreasing).

Any feedback would be appreciated!

varaunight
  • 21
  • 3

2 Answers2

2

Consider the sequence $$a_1 = 1000001\\a_n = 1, \quad n> 1$$ The limit of this sequence is quite obviously $1$.

What about the sequence of averages $A_n =\dfrac{\sum_{i=1}^n a_i}n$? The first several values are well away from that limit: $$A_1 = 1000001\\A_2 = 500001\\A_3 = 333334.\overline 3\\\vdots$$

But that doesn't keep up forever. $$A_{1000000} = 2\\A_{10000000} = 1.1\\A_{100000000} = 1.01$$

As $n$ get larger, the impact of that extra initial $1000000$ is diminished, while the slow plodding accumulation of additional $1$s keeps up with growth of the denominator, keeping the contribution of the rest near $1$.

The same thing happens with any convergent sequence. It may incredibly volatile at the start, with insanely large numbers. It may even keep up that volatility for an insanely long time. But to say a sequence converges, that volatility has to come to an end. "Convergence" says that at some point the sequence becomes boring, forever afterward providing values that barely differ from its limit $A$. No matter how volatile the sequence was at the beginning, the index $n$ eventually becomes big enough that it dwarfs the sum $S_\text v$ of all the contributions of that volatile beginning. $\frac {S_\text v}n \approx 0$ for large enough $n$. But the plodding addition of values $\approx A$ once the volatility dies down is something $n$ can never outpace. The sum increases by about $A\Delta n$ when $n$ increases by $\Delta n$. The average remains about $A$, and as $n$ increases, this approximation improves.

Paul Sinclair
  • 45,932
0

$\{a_i\}$ converges to $A$

That is, there exists some $N$ such that $n>N \implies |a_n-A| < \epsilon$

We can now break up the sequence into a finite portion where the index is less than or equal to $N$, and an infinite sequence that is arbitrarily close to $A.$

$\lim_\limits{n\to \infty} \frac 1n \sum_\limits{i=0}^N {a_i} + \frac 1n \sum_\limits{i=N+1}^n a_i$

As $n$ goes to infinty the first term goes to 0, and the second term approaches $A$

user317176
  • 12,247