Let $\{ a_n\}$, $\{ b_n\}$ be two sequences where $b_n$ is increasing such that $ \lim_{n \rightarrow \infty}b_n = \infty$. Also that $$\lim_{n \rightarrow \infty} \frac{a_{n+1}-a_n}{b_{n+1}-b_n} = s$$ Show that $\lim_{n \rightarrow \infty}\frac{a_n}{b_n} = s$. Further show that if $b_n >0$ then we have $$\lim_{n \rightarrow \infty} \frac{a_1+a_2+ . . . . . +a_n}{b_1+b_2+ . . . . . +b_n} = s.$$

I proceeded $$ \lim_{n \rightarrow \infty} \{(a_{n+1}-a_n) - s(b_{n+1}-b_n)\} = 0$$ Then dividing this by $b_n$ gives me $$ \lim_{n \rightarrow \infty}\{s - \frac{a_n}{b_n}\} + \lim_{n \rightarrow \infty}\{ \frac{a_{n+1}}{b_n} - s\frac{b_{n+1}}{b_n} \} = 0$$ Proceeding in a similar fashion with $$\lim_{n \rightarrow \infty} \frac{a_{n}-a_{n-1}}{b_{n}-b_{n-1}} = s$$ I got $$\lim_{n \rightarrow \infty} \frac{a_{n+1}-a_{n-1}}{b_{n+1}-b_{n-1}} = s$$

  • $\begingroup$ Is this what you have done so far? Or a complete proof? $\endgroup$ – ajotatxe Jul 12 '15 at 9:45
  • $\begingroup$ Yes I have done this much. I am still trying. I'll edit if I proceed further $\endgroup$ – silentkiller Jul 12 '15 at 9:48
  • $\begingroup$ For you reference: imomath.com/index.php?options=686 $\endgroup$ – Siminore Jul 12 '15 at 9:59
  • $\begingroup$ Your first limit after "I proceeded" does not need to exist. For example, take $a_n=n^2$, $b_n=n^3$, $s=0$. $\endgroup$ – celtschk Jul 12 '15 at 10:11

This is the Stolz-Cesaro Theorem and you can google it.

Highlights of the proof: for any arbitrary $\;\epsilon>0\;$ :

There exists $\;N\in\Bbb N\;$ s.t. for $\;n>N\;$ :


Multiply through by $\;b_{n+1}-b_n\;$ and sum up from $\;i=N\;$ to $\;i= k>N\;$ :


Observe we have here telescopic series...Then, divide by $\;b_{k+1}\;$:

$$(l-\epsilon)\left(1-\frac{b_N}{b_{{k+1}}}\right)+\frac{a_{N}}{b _{{k+1}}}\le\frac{a_{k+1}}{b_{k+1}}\le(l+\epsilon)\left(1-\frac{b_N}{ b_{k+1}}\right)+\frac{a_N}{b_{{k+1}}}$$

End now the argument.

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