# Let $(x_n)$ be a bounded sequence such that $x_{n+1} \geqslant x_n - \frac{1}{2^{n}}$ for every $n \in \mathbb{N}$, show that $(x_n)$ converges.

Let $(x_n)$ be a bounded sequence such that $x_{n+1} \geqslant x_n - \frac{1}{2^{n}}$ for every $n \in \mathbb{N}$, show that $(x_n)$ converges.

I know that if I can prove that $(x_n)$ is monotone, then by the monotone convergence theorem, $(x_n)$ would converge, but I don't know where to start to show that. There was a suggestion saying that I should first prove that $x_n - \frac{1}{2^{n}}$ is monotonically increasing, but that leads to a contradiction.

• What contradiction was that? Proving that $x_n - \frac{1}{2^{n}}$ is monotonically increasing seems the best way to prove what you want – wilkersmon Feb 21 '18 at 19:00
• @Math_QED the series for $\ln(2)=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}$ is a bounded convergent sequence but it's not monotone, so your statement is false. – wilkersmon Feb 21 '18 at 19:01
• Sorry mixed up two things. Comment deleted. – user370967 Feb 21 '18 at 19:07
• Why would it converge?? If $x_{n+1}$ can be greater than $x_n$, then its enough to pick $x_0 = 1$ to see that you are going to have an infinite sum of 1's or more. – Aleksejs Fomins Feb 21 '18 at 19:48
• @AleksejsFomins: there is no series here, just a sequence. – Jason DeVito Feb 21 '18 at 19:57

Put $$y_n:=x_n-{1\over 2^{n-1}}\leq\sup_k x_k=\xi<\infty\ .$$ Then $$y_{n+1}=x_{n+1}-{1\over 2^n}\geq x_n-{2\over 2^n}=y_n\qquad(n\geq1)\ .$$ It follows that the $y_n$ form a bounded increasing sequence, hence converge to a real number $\eta$. This allows to conclude that $$x_n=y_n+{1\over 2^{n-1}}\to\eta\qquad(n\to\infty)\ .$$