# Given a bounded above sequence such that $x_{n+1}-x_n \ge -{1\over 2^n}$, prove that $x_n$ converges. [duplicate]

Prove that a bounded above sequence converges given it satisfies the following property: $$x_{n+1}-x_n \ge -{1\over 2^n}\\ n\in\Bbb N$$

Since the sequence is bounded above, by definition we have: $$\exists M \in\Bbb R: x_n \le M, \forall n\in\Bbb N$$

By reversing the sign: $$x_{n+1} - x_{n} \ge -{1\over 2^n} \iff x_{n} - x_{n+1} \le {1\over 2^n}$$

Then I tried to consider a list of inequalities: $$x_1 - x_2 \le {1\over 2^1}\\ x_2 - x_3 \le {1\over 2^2}\\ x_3 - x_4 \le {1\over 2^3}\\ \cdots\\ x_{n} - x_{n+1} \le {1\over 2^n}$$

If we now sum up the inequalities one may obtain: $$x_1 - x_2 + x_2 - x_3 + x_3 - x_4 + \cdots + x_{n} - x_{n+1} \le \sum_{k=1}^n{1\over 2^k}$$

By telescoping we obtain: $$x_1 - x_{n+1} \le \sum_{k=1}^n{1\over 2^k}$$

So: $$x_1 - x_{n+1} \le 1 - {1\over 2^n} \iff \\ M \ge x_{n+1} \ge x_1 - 1 + {1\over 2^n}$$

I'm not sure about where to go from here. How to prove what's in the problem statement?

## marked as duplicate by Martin R, TheSilverDoe, Community♦Mar 21 at 15:00

• $$x_n = x_1 - \underbrace{\sum_{k=1}^{n-1}(x_k- x_{k+1})}_{convergent}$$
It follows the convergence of $$(x_n)$$