Convergence of $b_n = \frac{\sum_{i=1}^na_i}{4}$

Question:

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 $$\{b_n\}$$.

Attempt:

Since $$\{a_n\}$$ converges to $$A$$, for each $$\epsilon>0$$, there exists $$N>0$$ such that whenever $$n\geq N$$, $$A-\epsilon. We can write $$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}

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 :(

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