Is this a valid proof to prove that if $a_n$ converges, then $a_{n+1}-a_{n}$ converges to $0$? by definition I saw someone's comment on another website on this proof and they presented this (by definition):  

now, by the definition of a convergent sequence,
  for all $\epsilon _1 >0$ there is in fact an $n> N_1$ such that it ensures $|a_n - L| < \epsilon _1$, where $L$ is the limit of $a_n$.
  Now, this means that $a_{n+1}$ is also a convergent sequence so
  for all $\epsilon _2 > 0$ there is in fact an $n> N_2$ such that it ensures $|a_{n+1} - L| < \epsilon_2$, where $L$ is the limit of $a_{n+1}$.
  Now consider,
  $|a_{n+1} - a_{n}| < \epsilon$ for all $\epsilon > 0$ for some $n>N$
  Now if we were to pick $N = \max\{N_1,N_2\}$, then we have our result.


Yes. This is correct. Using the triangular inequality: $|a_{n+1} - a_n| = |a_{n+1} -L + L - a_n| \le |a_{n+1} - L| + |a_n - L| < \epsilon$, therefore when you take the limit you get $a_{n+1} - a_n = 0$.
