Let $\{a_n\}_{n\in \mathbb N}$ be a sequence of real numbers, and for each $n\in \mathbb N$ define $$b_n= \frac{\sum_{i=1}^na_i}{4}.$$ Prove that if $\{a_n\}$ converges to $A$, then so does $\{b_n\}$.


Since $\{a_n\}$ converges to $A$, for each $\epsilon>0$, there exists $N>0$ such that whenever $n\geq N$, $A-\epsilon<a_n<A+\epsilon$. We can write $b_n$ as $$ b_n=\frac14\left(\sum_{i=1}^Na_i + \sum_{i=N+1}^na_i\right),$$ and by the convergence of $\{a_n\}$, we can write $$(n-N)(A-\epsilon)<\sum_{i=N+1}^na_i<(n-N)(A+\epsilon).$$ If we write $C=\sum_{i=1}^Na_i$, then $$\frac{C+(n-N)(A-\epsilon)}{4}<b_n<\frac{C+(n-N)(A+\epsilon)}{4}.$$

We did an example where $b_n$ was the arithmetic mean and it worked because the denominator was $n$ and we could see that in the large $n$ limit, $|b_n-A|<\epsilon$. But here, I'm not sure how to take it home, because $(n-N)$ can grow without bound. Am I even on the right path?

Edit: The proposition is false :(

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    $\begingroup$ Take $\{a_n\}$ to be the constant sequence $A$. $b_n = \cfrac{nA}{4} \to +\infty$ $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 9 '18 at 19:58
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    $\begingroup$ Perhaps there is a typo, and this math.stackexchange.com/q/155839/42969 is what you really mean? $\endgroup$ – Martin R Dec 9 '18 at 20:00
  • $\begingroup$ @MartinR which is solved right away with Stolz-Ces$\mathrm{\grave{a}}$ro !!!!. $\endgroup$ – Felix Marin Dec 9 '18 at 21:03
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    $\begingroup$ @WyattKuehster: Then your statement is obviously wrong, as pointed out above. More generally, if $a_n \to A > 0$ then $b_n \to \infty$. $\endgroup$ – Martin R Dec 9 '18 at 22:26
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    $\begingroup$ Or take any non-negative sequence such that $\sum a_n$ diverges, such as $a_n = 1/n$. $\endgroup$ – Martin R Dec 9 '18 at 22:34

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