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We are given that $(s_{n})$ is a bounded sequence. My goal is to show that $(s_{n})$ is monotonic, and thus it converges. It has the following property $$s_{n+1} \geq s_{n} - \frac{1}{2^n}$$ I have tried induction, because intuitively the sequence monotonically increasing, but it fails at the basis step but works on the induction step. Any ideas?

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That doesn't look like a definition but rather like a property of the sequence... – DonAntonio Mar 3 '14 at 11:02
True, edited now. – user29163 Mar 3 '14 at 11:05
Do you need to prove monotonicity or you need only to prove the sequence converges? – DonAntonio Mar 3 '14 at 11:14
Thanks, I only need to prove that the sequence converges. I since it is bounded I assumed (big mistake) that the right move is to try to prove monotonicity. I will try other things. Learned much from this case. Thanks – user29163 Mar 3 '14 at 11:22
up vote 2 down vote accepted

If you need to prove convergence:

$$s_n-s_{n+1}\le\frac1{2^n}\implies \;\forall\,m>n\;:$$



So our sequence is a Cauchy sequence and thus it converges (one can see this without the last step above as the left expression in the last line is the difference $\;S_n-S_m\;$ of the sequence of partial sums of the convergent series $\;\sum\frac1{2^n}\;$...)

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Counter-example. Bounded but non-monotonic. $s_{n+1}$ = \begin{cases} 0, & n+1 \mbox{ is even, }\\ \frac{-1}{2^n}, & n+1 \mbox{ is odd.} \end{cases}

E.g. $s_1 = 0, s_2 = -0.5, s_3 = 0, ....$

$s_1 > s_2$ but $s_2 < s_3$.

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$s_n$ is not necessarily monotonic given the information provided.

Example. Let $s_n = \frac{2}{3} \left(\frac{-1}{2}\right)^n $. Then $s_n$ is bounded, and \begin{align*} s_{n+1} - s_n &= \frac{2}{3}\left(\frac{-1}{2}\right)^{n+1} - \frac{2}{3}\left(\frac{-1}{2}\right)^n \\ &= \frac{2}{3}\left(\frac{-1}{2}\right)^n \left( \frac{-1}{2} - 1\right) \\ &= -\left(\frac{-1}{2}\right)^n \\ &= (-1)^{n+1} \left(\frac{1}{2}\right)^n \ge \frac{-1}{2^n}. \\ \end{align*} However, $s_n$ is clearly not monotonic.

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