Prove Sn is a Cauchy sequence and hence a convergent sequence.

Question

I am learning real analysis and I am stuck on this one proof. It reads:

Let $S_{n}$ be a sequence such that $|S_{n+1}-S_{n}| < 2^{-n}$

Prove $S_{n}$ is a Cauchy sequence and hence a convergent sequence.

Here is what I've tried doing:

Proof:

Let $\epsilon > 2^{-n}$, we want to show that $\exists$ $N$ such that $n+1, n$ $>$ $N$ implies $|S_{n+1}-S_{n}| < 2^{-n}$

Using the triangle inequality we get, $-2^{-n}< S_{n+1}-S_{n} <2^{-n}$

Further simplification, $2^{-n}+S_{n+1} > S_{n}> -2^{-n}+S_{n+1}$

This is where I got up to and I feel as thought I am on the wrong track.

Answer 1

Let $m > n$. Then $|S_m - S_n| \leq \sum_{k=n+1}^m |S_{k+1}-S_k| \leq \sum_{k=n+1}^{\infty} 2^{-k} = 2^{-n}$. So given any $\epsilon$, pick $n$ such that $2^{-n} < \epsilon$ and $m > n$ to satisfy the Cauchy criterion.

Answer 2

Use the definition. WLOG let $N<n\le m$ then

$$|S_m-S_n|\le |S_m-S_{m-1}|+|S_{m-1}-S_{m-2}|+\ldots + |S_{n+1}-S_n|$$ $$ 2^{-m-1}+\ldots + 2^{-n}\le 2^{-m}\le 2^{-N}$$

So for $0< \epsilon <1$ choose $N> -\log_2(\epsilon)$.

Answer 3

You conception of a Cauchy sequence is wrong: you have to prove that, given $\varepsilon>0$, there exists a rank $N$ such that $\lvert S_n-S_p\rvert<\varepsilon$ for any $n,p> N$.

Answer 4

Hint

For $m>n$ we have $$|s_m - s_n| = \left| s_m -s_n + \sum_{j=n+1}^{m-1} (s_j -s_j) \right|= \left| \sum_{j=n}^{m} (s_{j+1} -s_j)\right| \leq \sum_{j=n}^{m} \frac{1}{2^j}.$$

What do you know about the remainder of a convergent series?