Prove the sequence converges We have a bounded sequence $a_n$ such that, $a_n - a_{n+1} \le 1/2^n$.
Define $b_n = a_n - 1/2^{n-1}$
I need to show $a_n$ and $b_n$ converge.
For me, The intuition Is that since the difference between two $a_n$ terms is very small when ${n\rightarrow \infty}$, $a_n$ must converge, But I have difficulties proving it.
Plus, I don't understand why they are asking calculate the limits of both $b_n$ and $a_n$. They should be equal!

If you have $|a_n-a_{n+1}|<2^{n-1}$, then you can use the standard trick to show that $\{a_n\}$ is a Cauchy sequence by show that $|a_m-a_n|<=|a_m-a_{m+1}|+\cdots+|a_{n-1}-a_n|<1/2^m+\cdots<1/2^{m-1}$ which is sufficiently small for sufficiently large $m$, but the absolute value is missing here.
To overcome this difficulty, we need to use the condition that $a_n$ is bounded with the help of the sequence $\{b_n\}$. The condition implies that $b_n-b_{n+1}<=0$, therefore $\{b_n\}$ is decreasing and it's easy to show that $b_n$ is bounded from $a_n$ is bounded. So $b_n$ converges, and $a_n$ converges follows.
