Consider two sequences $A_n$ and $B_n$ such that $B_n$ is monotonically decreasing and both $A_n$ and $B_n$ tend to zero.

Now let us consider the limits $\lim_{n\to\infty}\left(\frac{A_n}{B_n}\right)$ and $\lim_{n\to\infty}\left(\frac{A_n-A_{n-1}}{B_n-B_{n-1}}\right)$. I guess that these two limits are equal when n tends to infinity if the second limit exists .

I tried to prove this using the following logic :

For any convergent sequence $C_n$, tending to limit $K$, $\lim_{n\to\infty}(C_n-k)= \lim_{n\to\infty}(C_n-C_{n-1})$. So as $A_n$ and $B_n$ tend to zero so $\lim_{n\to\infty}(A_n-0)= \lim_{n\to\infty} (A_n-A_{n-1})$ and the same for $B_n$. Hence if the limit second limit exists the first limit exists and is equal to the second limit.

But my stance is pretty skeptic and I feel an absence of rigor and strong reasoning.

Hope the members here can help me in strengthening the logic of this proof.


This is much like the Stolz-Cesaro theorem:

Consider sequences $\{a_n\}_{n=1}^\infty,\{b_n\}_{n=1}^\infty$ such that $b_n$ is strictly increasing and unbounded.

If $\lim\limits_{n\to\infty}\left(\frac{a_n-a_{n-1}}{b_n-b_{n-1}}\right)=l $ exists, then $\lim\limits_{n\to\infty}\dfrac{a_n}{b_n}$ exists and is equal to $l$.

A proof can be found here http://www.imomath.com/index.php?options=686

  • $\begingroup$ @Americo I still find this proof havin sort of gaps..Like the sixth line... could you explain or elaborate? $\endgroup$ – danny gotze May 16 '13 at 17:56
  • $\begingroup$ can i know where i was wrong? $\endgroup$ – danny gotze May 16 '13 at 18:05

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