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Suppose {Sn} has the property that abs(Sn+1 - Sn)<=2^-n for all n in N. Prove {Sn} is Cauchy.

I want to use the definition: A sequence {Sn} is Cauchy if for every e>0, there is a natural number N s.t. for all m,n>=N we have abs(Sm-Sn)<e.

So, I assume {Sn} is Cauchy and let e>0 be given but I don't know where to go from here.

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You cannot assume $(S_n)$ is Cauchy: that is what you have to prove. You have to show that, given any $\varepsilon > 0$, you can find $N=N(\varepsilon)$ such that the condition you have written is satisfied for all $m,n\geq M$.

Hint: Assuming without loss of generality that $m \geq n$ $$\begin{align} \lvert S_m - S_n\rvert &= \lvert S_m - S_{m-1}+S_{m-1}-S_{m-2}+\dots + S_{n+1} - S_n\rvert \\ &\leq \sum_{k=n}^{m-1} \lvert S_{k+1} - S_k\rvert \leq \sum_{k=n}^{m-1} \frac{1}{2^k} \leq \sum_{k=n}^{\infty} \frac{1}{2^k} \\ &= \frac{1}{2^{n-1}} \end{align} $$

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Given $\epsilon > 0$, suppose $m > n$ and write $m = n + k$. Now,

$$|S_{n+k}-S_n| = |S_{n+k}-S_{n+k-1} + \ldots + S_{n+1} - S_{n}|$$

But then, by the triangle inequality,

$$|S_{n+k}-S_n| \leq |S_{n+k}-S_{n+k-1}| + \ldots + |S_{n+1} - S_{n}|$$

But then,

$$|S_{n+k}-S_n| \leq 2^{-(n+k-1)} + \ldots + 2^{-n}$$

The right hand side gives the sequence of partial sums of the convergent geometric series

$$\sum_{i=n}^{\infty} 2^{-i},$$

so the write hand side is a convergent sequence, and thus it is Cauchy. This means that, for a sufficienty large $n$, the right hand side will be smaller than $\epsilon$, and this proves that $S_{n}$ is Cauchy.

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