At which step of the proof below I am making a mistake? $(a_n) \rightarrow 0 \implies \sum^\infty_{n=1} a_n \rightarrow L \;\;$ for some $L < \infty$
I know the above statement is false. A typical counterexample is  $a_n = \frac{1}{n} \;\;$. Yet, when I attempt to prove somehow I "prove" it. So I am doing a mistake in my proof but I couldn't figure out where. I was hoping someone can point out the mistake.
given $\forall \epsilon > 0, \; \exists N : \forall n > N, \; |a_n| < \epsilon$
I want to show that for $\;\; s_n = a_1 + ... + a_n \;\; ,\;\;\;\forall \epsilon > 0, \;\; |s_n - s_m| < \epsilon $ 
Without loss of generality assume that $n>m \;$ and $\;n = m+k\;$, then $\;\; |s_n - s_m| = |a_n + ... + a_{m+1}|$
Now, since $(a_n) \rightarrow 0 \;\;$, given $\epsilon >0\;$ pick $N^*>0$ such that  $\forall m> N^*, |a_m| < \frac{\epsilon}{k}$.
then let $n,m >N^*$:
$|s_n-s_m|  = |a_{m+k} + ... + a_{m+1}| \leq |a_{m+k} |+  ... + |a_{m+1}| \leq \frac{\epsilon}{k} + ... + \frac{\epsilon}{k} = k \frac{\epsilon}{k} = \epsilon $
I am depending the choice of $N^*$ also on the value $k(n,m)$. So if I pick another $n' > m+k$ then the above inequality fails. So this means the above argument only holds given $\forall m, m+k > N^* \;\;$ and $\forall k >0$? I am confused a bit of help on pointing out my mistake will be appreciated...
 A: Be careful in the proof. You have to show the following:
For all $\epsilon > 0$ there is $N^*$ large enough, such that for all $n,m > N^*$, we have $|s_n - s_m| < \epsilon$.
First of all : WLOG, $n = m+k$, where $k$ isn't fixed, it can be any natural number.
You are saying the following : for the given $\epsilon$, there is $N^*$ such that $|a_m| < \frac{\epsilon}{k}$. This certainly cannot hold for all $k$, since then you are saying that $a_m = 0$ after $N^*$, which need not be the case  for example in the harmonic series. Hence, $k$ is actually fixed by this statement.
Now, $n,m > N^*$, then provided $n=m+k$ (remember, $k$ is fixed!), we have that $|s_n-s_m| < \epsilon$.
However, the question arises : we need the equality for all $n,m$ which isn't true,  this is being provided only when the distance between them is fixed! 
Hence, the argument is flawed.
To give an example: suppose we take the Harmonic series $\frac 1n$, and $ \epsilon = \frac 19$. Then, yes, we can find, for all $k$, a number $N^*$ depending on $k$, such that after $N^*$, the sum of $k$ consecutive terms is bounded by $\epsilon$. However, we can't say much if we are considering the sum of $1000k$ terms after $N^*$, for example.
