Where are the errors in this convergence proof? The following proof is wrong, and I'm not 100% sure why.

$a_n$ is a sequence s.t. $|a_{n+1} - a_{n}| < \frac{1}{n}$. Determine whether $a_n$ converges.
We want to show that $\forall \epsilon > 0, \exists N: n, m > N \implies |a_m - a_n| < \epsilon$.
WLOG, let $m \geq n$.
Let $k = m - n$, so that $n + k = m$.
Note that $\lim |a_{n+1} - a_{n}| = 0$.
$|a_m - a_n|$
$= |a_{n+k} - a_n|$
$= |(a_{n+k} - a_{n+k-1}) + (a_{n+k-1} - a_{n+k-2}) + \ldots + (a_{n+1} - a_n)|$
$\leq |a_{n+k} - a_{n+k-1}| + |a_{n+k-1} - a_{n+k-2}| + \ldots + |a_{n+1} - a_n|$
Therefore:
$\lim |a_{m} - a_n|$
$\leq \lim(|a_{n+k} - a_{n+k-1}| + |a_{n+k-1} - a_{n+k-2}| + \ldots + |a_{n+1} - a_n|)$ (1)
$= \lim |a_{n+k} - a_{n+k-1}| + \lim |a_{n+k-1} - a_{n+k-2}| + \ldots + \lim |a_{n+1} - a_n|$ (2)
$= 0 + 0  + \ldots + 0$ (3)
$= 0$
So $|a_{m} - a_n|$ converges to 0, meaning $a_n$ is Cauchy, and hence converges.

My guess is, this proof doesn't work because you can't always go from (1) to (2), but (2) to (3) could also be suspect.
I'd like to hear thoughts from a more experienced mathematician.
 A: First, note that $$\log(n+1)-\log(n) = \log(1+\frac{1}{n}) < \frac{1}{n},$$ so it isn't just that the proof is wrong: the statement isn't true.
The problem with the proof is with your order of limits. To show it's Cauchy, you need to show for all $\epsilon>0$ there is an $N$ such that for any $m,n >N,$ $|a_m-a_n|<\epsilon.$ In your argument, you show (indirectly) that you can find an $N$ for a given $k$ such that $m=n+k,n,$ but you don't show that you can find one $N$ that works for all $n,m.$
In order to iterate the small gap between $a_{n+1}$ and $a_n$ into a small gap between $a_{n+k} $ and $a_n$ you need to add up $k$ gaps, all of which go like $\frac{1}{n+k}.$ Since $\sum_{k} \frac{1}{n+k}$ does not converge, there is no guarantee that the size of the gap $|a_{n+k}-a_n|$ can be controlled for arbitrarily large $k$. In other words, $$ \infty = \lim_{n\to \infty} \sum_{k=1}^\infty \frac{1}{n+k} \ne \sum_{k=1}^\infty \lim_{n\to\infty}\frac{1}{n+k} = 0.$$
A: Your error is the equals-sign at the beginning of line (2). You're presumably thinking of the principle "the limit of a sum equals the sum of the limits (provided the latter exists)." But this principle is correct only when the number of summands is fixed.  In your situation, the number $k$ of summands is not fixed, and so the principle does not apply.
For a simpler example of this situation, consider the sum $\frac1n+\frac1n+\cdots+\frac1n$, where there are $n$ summands. So this sum is simply $1$. The limit of the sum as $n\to\infty$ is therefore also $1$. Yet each of the summands $\frac1n$ approaches $0$ as $n\to\infty$.
A: HINT: Let $a_1=1$, $a_{n+1}=a_n+\dfrac{1}{n+1}$. Then $a_{n+1}-a_{n}=\dfrac1{n+1}$. 
Is such $(a_n)$ convergent?
What is (your) $|a_{n+k} - a_{n+k-1}| + |a_{n+k-1} - a_{n+k-2}| + \ldots + |a_{n+1} - a_n|$ for this sequence?
