Proof review: cauchy sequences Problem statement: 
a) Let $(s_n)$ be a sequence such that $\mid s_{n+1} - s_{n} \mid < 2^{-n}$ for all $n \in \mathbb{N}$. Prove that $(s_n)$ is a Cauchy sequence and hence a convergent sequence.
b) Is the result in (a) treu if we only assume that $\mid s_{n+1} - s_{n} \mid < \frac{1}{n}$ for all $n \in \mathbb{N}$?
My proof for (a):
Take any $\epsilon > 0$. Let $N = \log_2{2/\epsilon}$. Take $m, n > N$. Without loss of generality, we may take $m > n$. Let $\delta = m-n$. 
$\mid s_{n+1} - s_{n} \mid < \frac{1}{2^n}$
$\mid s_{n+2} - s_{n+1} \mid < \frac{1}{2^{n+1}}$
...
$\mid s_{n+\delta} - s_{n+\delta-1} \mid < \displaystyle{\frac{1}{2^{n+\delta-1}}}$
Adding up the inequalities and using the triangle inequality:
$\mid s_{n+\delta} - s_n \mid = \mid s_{m} - s_{n} \mid < \displaystyle{\frac{1}{2^{n}} + \frac{1}{2^{n+1}} ... \frac{1}{2^{n+\delta-1}}}$
Using the geometric series sum formula:
$\displaystyle{\frac{1}{2^{n}} + \frac{1}{2^{n+1}} ... \frac{1}{2^{n+\delta-1}}} = \frac{1}{2^{n-1}} - \frac{1}{2^{n+\delta-1}} < \frac{1}{2^{n-1}} - \frac{1}{2^n} < \frac{1}{2^n-1} < \frac{1}{2^{N-1}} = \epsilon$
Therefore,
$\mid s_m - s_n \mid < \epsilon$.
For (b), it doesn't work because the argument is that $\sigma \frac{1}{n}$ does not converge, but I'm not sure how to flesh that out. 

For part (b), try this counter-example:
$$s_n = \sum_{k=1}^{n} \frac{1}{n}.$$
Then
$$|s_{n+1} - s_{n}| = \frac{1}{n+1} < \frac{1}{n},$$
but $s_n \to \infty$.
