Prove a sequence of real numbers is a Cauchy Sequence 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.
 A: 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 $m \geq n$
$$\begin{align}
\lvert S_m - S_n\rvert &= \lvert S_m - S_{m-1}+S_{m-1}-S_{m-2}+\dots + S_{n+1} - S_n\rvert \\
&\leq \sum_{k=n}^{m-1} \lvert S_{k+1} - S_k\rvert 
\leq \sum_{k=n}^{m-1} \frac{1}{2^k} 
\leq \sum_{k=n}^{\infty} \frac{1}{2^k} \\
&= \frac{1}{2^{n-1}}
\end{align}
$$
A: 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.
