How to show that this sequence is monotonic 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? 
 A: $s_n$ 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.
A: If you need to prove convergence:
$$s_n-s_{n+1}\le\frac1{2^n}\implies \;\forall\,m>n\;:$$
$$s_n-s_m=(s_n-s_{n+1})+(s_{n+1}-s_{n+2})+\ldots+(s_{m-1}-s_m)\le$$
$$\le\frac1{2^n}+\frac1{2^{n+1}}+\ldots+\frac1{2^{m-1}}\le\frac{m-n-1}{2^n}$$
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 $\;\sum\frac1{2^n}\;$...)
A: 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 $s_2 < s_3$.
