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I apologize in advance for the poor formatting of this question - I'm in a coffee shop, on my phone since campus buildings closed early today..

Given:

$\{a_n\}$ and $\{b_n\}$ are both positive sequences.

The infinite series of $\{b_n\}$ diverges to infinity.

$\frac{a_n}{b_n} \rightarrow c$ for $c$ nonnegative.

Define $A_n = a_1 + ... + a_n$ and $B_n = b_1+b_2+...+b_n$, for $n ≥ 1$. Show that $\frac{A_n}{B_n} \rightarrow c$.

I wrote out $\frac{a_1+...+a_n}{b_1+...+b_n}$ as

$$\frac{a_n( \frac{a_1}{a_n} + ... + \frac{a_{n-1}}{a_n} + 1)}{ b_n (\frac{b_1}{b_n} + ... + \frac{b_{n-1}}{b_n} + 1)}$$

So I just factored out an $a_n$ from the top and a $b_n$ from the bottom since I want to use the fact that $\frac{a_n}{b_n} \rightarrow c$. I need to somehow show the other factor converges to $1$ as n goes to infinity. I probably need to use the fact that series $b_n$ diverges to infinity, but the $+1$ on the top and bottom have made things tricky so far.

If I can do this, then I will have shown that $\frac{A_n}{B_n}$ also converges to $c$.

The student solution that I took a look at is probably wrong, as it assumes that $b_n$ diverges to infinity but that is not necessarily true, eg, series $\frac{1}{n}$ diverges to infinity, but the sequence $\frac{1}{n}$ converges.

Thanks in advance,

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    $\begingroup$ BTW the claim is called Stolz-Cesaro theorem. (Or Stolz theorem in some books.) $\endgroup$ – Martin Sleziak Jan 1 '15 at 11:04
  • $\begingroup$ I studied the proof of this theorem from the IMO page and can follow most of it by now, @martinsleziak. What do you you think of the proof? (The one that proves the string of inequalities, liminf an/bn <= liminf An/Bn <= limsup An/Bn <= limsup an/bn.) $\endgroup$ – User001 Jan 2 '15 at 0:19
  • $\begingroup$ If you mean the proof here, it seems to be more or less standard proof. (I prefer this proof written for sums, but it is probably a matter of taste.) $\endgroup$ – Martin Sleziak Jan 2 '15 at 7:53
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$$\frac{A_n}{B_n}-c=\frac{\sum_{k=1}^n (a_k-cb_k)}{B_n}$$ Since $\dfrac{a_n}{b_n}\to c$, $\forall \epsilon>0,\ \exists N\in \mathbb{Z}^+$ such that $\forall n\ge N$ $$|a_n/b_n-c|\le \epsilon$$ Then, we have, $$|\frac{A_n}{B_n}-c|\le \frac{|\sum_{k=1}^{N-1}(a_k-cb_k)|}{B_n}+\epsilon\frac{\sum_{k=N}^n b_k}{B_n}\le \frac{|\sum_{k=1}^{N-1}(a_k-cb_k)|}{B_n}+\epsilon $$, now since $B_n$ is a positive increasing sequence and $B_n$ diverges, for any given $\delta>\epsilon>0$ one can find $N_1\ge N$ such that $\forall n\ge N_1$ $$\frac{|\sum_{k=1}^{N-1}(a_k-cb_k)|}{B_n}\le \delta-\epsilon\Rightarrow |\frac{A_n}{B_n}-c|\le \delta$$ Hence $\dfrac{A_n}{B_n}\to c$

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  • $\begingroup$ Thanks so much,@ samratmukhopadhyay. It turns out that this question is a general form of the Stolz-Cesaro theorem, if you're interested. $\endgroup$ – User001 Jan 1 '15 at 1:50
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    $\begingroup$ Oh, nice. I know Cezaro's theorem but I did not know its generalization. Thanks for the reference @LebronJames :-) $\endgroup$ – Samrat Mukhopadhyay Jan 1 '15 at 10:02

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