Where is the mistake? Insufficient condition for convergence. I would like to solve the following problem

My attempted solution is as follows. Convergence implies that given $\epsilon>0$, the
r exists $N$ such that for all $n\geq N$, we have
\begin{align*}
|a_{n+k}-{a}_{n} |<\epsilon.
\end{align*}
We are told that this has to hold for all $k$. Consider any $\alpha\geq \beta \geq N
$. Then , by the above, we can find $N$ such that
\begin{align*}
 |a_{\beta +k}-a_{\beta }|<\epsilon
 \end{align*}
for all positive $k$. Now, as $\alpha \geq \beta $, we can set $k=\alpha -\beta $ (in
the case $\alpha =\beta $ we have $|a_{\alpha }-a_{\beta }|=0<\epsilon$) giving us
\begin{align*}
|a_{\alpha }-a_{\beta }|<\epsilon.
\end{align*}
So given $\epsilon>0$, there exists $N$ such that if $\alpha ,\beta >N$, then
\begin{align*}
|a_{\alpha }-a_{\beta }|<\epsilon,
\end{align*}
meaning that the sequence $({a}_{n} )$ is Cauchy, and hence convergent.
However, this is wrong as per the counter example
\begin{align*}
{a}_{n} = \sum_{i=1}^{n} \frac{1}{i}.
\end{align*}
Can someone help me find the mistake?
 A: The claim is incorrect for if $a_n=\log (n)$ then for any fixed $k$ we have
$$a_{n+k}-a_n = \log(n+k)-\log(n) = \log\left(1+\frac{k}{n}\right)$$
which tends to $0$ as $n\to\infty$.
The mistake in your attempted proof is in the very first line. You write that for every $\epsilon>0$ there exists $N$ such that for all $n>N$ we have
$$a_{n+k}-a_n<\epsilon$$
but since $k$ appears in this formula we need to be careful about what this exactly means. Do you mean that for every $k$ and for every $\epsilon$ there exists such $N$? If so, the claim is correct. However, if you say that for every $\epsilon>0$ there exists $N$ which applies for every $k$, then the claim is wrong. Indeed, later on you use the fact that this $N$ supposedly works for every $k$, which is wrong, hence the proof fails.
Given $\epsilon>0$, for every $k$ we can find $N$. This doesn't mean there's a single $N$ which fits for every $k$. This is the distinction between
$$\forall\epsilon>0\,\forall k\,\exists N\,\ldots$$
and
$$\forall\epsilon>0\,\exists N\,\forall k\,\ldots$$
