If $|a_{n+1} - a_n| < 2^{-n}$ prove $(a_n)$ converges.

Question

The solutions present a different proof to what I came up with.

I am given $|a_{n+1} - a_n| < 2^{-n}$

Suppose $m$ >> $n$. Now : $$|a_{n+1} - a_m + a_m - a_n| < 2^{-n}$$ $$|a_{n+1} - a_m + a_m - a_n| \ge |a_m - a_n|-|a_{m}-a_{n+1}| \ge|a_m-a_n| < 2^{-n}$$

Now pick $N$ and $\epsilon > 0$ such that $2^{-N} < \epsilon$.

Then for $m > n > N$ I have $|a_m - a_n|< \epsilon \iff$ Cauchy therefore convergence.

Is this right? Or do I need to tweak some things

Answer 1

You proof is wrong because you can't conclude that $|a_m-a_n|<2^{-n}$, you can only say that $2^{-n}+|a_m-a_{n+1}|\geq |a_m-a_n|$. However this can be fixed, if $m> n$, applying your result repeatedly will lead you to $$2^{-n}+2^{-n+1}+\cdots+2^{-(m-1)} \geq |a_m-a_n|$$ from which you can easily show that $a_n$ is Cauchy.

Answer 2

Let $ n,p >0$.

$$|a_{n+p}-a_n|=$$ $$|a_{n+p}-a_{n+p-1}+...+a_{n+1}-a_n|\le$$

$$2^{-(n+p-1)}+...+2^{-(n+1)}+2^{-n}=$$

$$2^{-n}(1+2^{-1}+...+2^{-p+1})=$$

$$2^{1-n}(1-2^{-p})\le 2^{1-n}$$

for $ n $ great enough, we will have

$$|a_{n+p}-a_n|\le 2^{1-n}<\epsilon$$

$(a_n) $ is Cauchy, and converges.