Given a non-negative sequence $a_n$, strictly decreasing and tending to zero, can we show that (for large $n$)

$$ \frac{a_n - a_{n+1}}{a_n} \approx \frac{a_n}{na_n} = \frac{1}{n} \text{ }?$$

Obviously, we are inspired to attempt this by noticing that $a_n - a_{n+1} \to l \implies \frac{a_n}{n} \to l$, since the latter is (roughly) the Cesaro mean of the former (i.e., the Cesaro mean of the sequence $b_n = a_n - a_{n+1}$). If we had $l \neq 0$, then we could just note that the quotient of these two sequences would head towards 1, which would quickly give the result. However, I can't think of a way to handle this for the case where the sequence $a_n$ tends to zero.

Examples ($\frac{1}{n}, \frac{1}{\sqrt[3]{n}}, \frac{\log(n)}{n},$ etc.) suggest that the claim is true.

  • $\begingroup$ So your question is: for every non-negative decreasing sequence $(a_n)$, $(a_n-a_{n+1})/a_n\sim \frac{1}{n}$? $\endgroup$ – Hanul Jeon Nov 3 '14 at 5:13
  • $\begingroup$ There is a counterexample which converges to 0 much slower: consider $a_n=1/\ln(2+n)$. $\endgroup$ – Hanul Jeon Nov 3 '14 at 5:14
  • $\begingroup$ @HanulJeon you're right! There are many counterexamples to the claim. $\endgroup$ – user98186 Feb 12 '16 at 16:28

Try something that converges faster, like $a_n = 2^{-n}$.

Or more generally, for any $f$ with $0 < f(n) < 1$ and $c > 0$, you can define a sequence satisfying your conditions by

$$ a_0 = c \qquad \qquad a_{n+1} = a_n (1 - f(n)) $$

that has

$$ \frac{a_n - a_{n+1}}{a_n} = f(n) $$

  • $\begingroup$ Arrrgh. I don't know how I missed that. (Although I had a bad feeling about this plan from the beginning....) $\endgroup$ – Chris Nov 3 '14 at 5:17
  • $\begingroup$ Hurkyl, would you do me a favor and delete your answer, so I can delete this betise myself? $\endgroup$ – Chris Nov 3 '14 at 5:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.