# Showing that a sequence is monotone and thus convergent

Here's the original question:

Let $(a_n)$ be bounded. Assume that $a_{n+1} \ge a_{n} - 2^{-n}$. Show that $(a_n)$ is convergent.

Okay, I know that if I can show that if the sequence is monotone, I can conclude that it is convergent. But I am not sure how to show that it is monotone.

I know that $$a_n \le a_{n+1} + \frac{1}{2^n} < a_{n+1} + \frac{1}{n}$$

It looks to me as if it is monotonically increasing but I'm quite not sure how to prove my claim. Any hints would be appreciated.

For all $n$, let $$b_n = a_n - 2^{1-n}.$$ Note that $$b_{n+1} \ge b_n \iff a_{n+1} - 2^{-n} \ge a_n - 2^{1-n} \iff a_{n+1} \ge a_n - 2^{-n},$$ which is true. Note also that $b_n$ is the sum of the bounded sequence $a_n$ and the convergent sequence $-2^{1 - n}$, and hence is bounded as well. Thus, $b_n$ converges by the monotone convergence theorem, and hence, by the algebra of limits, so does $a_n$.
• How did you come up with $(b_n)$? – Ashish K May 27 '18 at 17:14
• I summed $$\sum_{k=0}^{n-1} 2^k = 2 - 2^{1-n}.$$ The idea is that the difference between $a_{n+1} - a_n$ and $b_{n+1} - b_n$ should be $2^{-n}$. Originally I used $b_n = a_n + 2 - 2^{1-n}$, but the constant addition of $2$ was superfluous. – Theo Bendit May 27 '18 at 17:25