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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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Now you write

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

It follows the convergence of $(x_n)$


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