Let $(a_n)_{n \in \mathbb{N}}$ and $(s_n)_{n \in \mathbb{N}}$ be sequences in $[0, \infty)$ so that $\sum\limits_{n=0}^{\infty}s_k$ converges and

$a_n \geq a_{n+1} -s_n ~~~~~~(n \in \mathbb{N})$

Show that $(a_n)_{n \in \mathbb{N}}$ is convergent.

A hint is given, i should sight the sequence $(b_n)_{n \in \mathbb{N}}$ with $b_n = a_n-\sum\limits_{n=0}^{n-1}s_k$ for $n \in \mathbb{N}$ first.

My idea is to show that the sequence is bounded above and monotonically nonincreasing. But how to do this?

  • $\begingroup$ Consider that $b_n$ is nonincreasing $\endgroup$ – Farshad Nahangi Dec 27 '13 at 21:57

We have $$b_{n+1}-b_n=a_{n+1}-a_n-s_n\le0$$ so the sequence $(b_n)$ is decreasing and $$b_n=a_n-\sum_{n=0}^{n-1}s_k\ge a_n-\sum_{n=0}^\infty s_k\ge-\sum_{n=0}^\infty s_k=C$$ hence $(b_n)$ is bounded below by $C$ so it's convergent. The convergence of $(a_n)$ can be deduced easily.

| cite | improve this answer | |
  • $\begingroup$ You have the nice skill of saying "enough" without saying "too much"! +1 $\endgroup$ – amWhy Dec 28 '13 at 15:52

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.