Is this a valid proof to prove that if $a_n$ converges, then $a_{n+1}-a_{n}$ converges to $0$? by definition

Question

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.

Answer 1

I don't find the proof particularly well-structured. While there are many ways to prove such a claim, if one wishes to stick close to the definition this might be a clearer approach:

Let $\varepsilon > 0$. For all $n$, we have $|a_{n+1} - a_n| = | a_{n+1} -L + L - a_n| \leq |a_{n+1} - L| + |L-a_n|$, by the triangle inequality. Because the sequence $(a_n)$ converges to $L$ by assumption, there is an $N_1$ such that for all $n > N_1$, we have $|a_n - L| < \frac{\varepsilon}{2}$, and similarly an $N_2$ so that $|a_{n+1} - L| < \frac{\varepsilon}{2}$. If we pick $N = \max \{N_1, N_2\}$, then for all $n > N$: $$|a_{n-1} - a_n| < 2 \cdot \frac{\varepsilon}{2} = \varepsilon.$$

What I find missing in the given proof is the use of the triangle inequality and what $\varepsilon_1$ and $\varepsilon_2$ are doing.

Answer 2

I guess that's fine, yet it'd be way easier and shorter if we use arithmetic of limits: since we're given $\;\{a_n\}\;$ converges, there exists $\;L\in \Bbb R\;$ s.t.

$$\lim_{n\to\infty}a_n=L\,.\;\text{ But trivially also}\;\;\lim_{n\to\infty}a_{n+1}=L\implies \;\text{by arithmetic of limits,}$$

$$\lim_{n\to\infty}(a_{n+1}-a_n)=\lim_{n\to\infty}a_{n+1}-\lim_{n\to\infty}a_n=L-L=0$$

and we're done.

Answer 3

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 $a_{n+1} - a_n = 0$.