Let $a_1,a_2,\ldots,a_{100}$ be real numbers,each less than one,satisfy $$a_1+a_2+\ldots+a_{100}>1$$ Prove the following statements:
(i) Let $n_0$ be the smallest integer $n$ such that $$a_1+a_2+\ldots+a_n>1$$
Show that all the sums $a_{n_0},a_{n_0}+a_{n_0-1},\ldots,a_{n_0}+\ldots+a_1$ are positive.
(ii) Show that there exist two integers $p$ and $q,p<q$, such that the numbers $$a_q,a_q+a_{q-1},\ldots,a_q+\ldots+a_p$$ $$a_p,a_p+a_{p+1},\ldots,a_p+\ldots+a_q$$
are all positive.

My work:
I could solve the first part by the Extremal Principle, but cannot approach the second part. I do not know extremal principal much. Please help!

  • $\begingroup$ @Arash...where did you edit? $\endgroup$ – Hawk Jan 27 '14 at 16:31
  • $\begingroup$ Just at the beginning where you put $a_100$. :) $\endgroup$ – Arash Jan 27 '14 at 16:36
  • $\begingroup$ but...didn't I rectify it? Anyways...must have missed it...thanks! $\endgroup$ – Hawk Jan 27 '14 at 16:37


Let $s_0$ be the largest integer $s$ (less than $n_0$) s.t.

$a_s + a_{s+1} + \dots + a_{n_0} > 1$

Then you can take $p=s_0$ and $q=n_0$ since if any of the sums in question is nonpositive it can be discarded to tighten the bounds (and the bounds were chosen "extremaly"). Also $p<q$ since the numbers $a_i$ are less than one.

| cite | improve this answer | |

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.