How to prove that $\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}$

Question

I want to prove the following theorem

Let $\{a_{n}\}$ be a convergent sequence. Prove that $\lim\limits_{n\to \infty}a_{n+1}=\lim\limits_{n\to \infty}a_{n}$

Do I need to show that it is monotone increasing and then use the relation

$$0\leq \lim_{n\to \infty}a_{n+1}- \lim_{n\to \infty}a_{n}\leq 0 \;?$$

Please, I need help on this! Various proofs are welcome! Thank you very much for your time!

Answer 1

Note that

• $|a_n-L|<\epsilon$ for $n>N\implies |a_{n+1}-L|<\epsilon$ for $n+1>N$

Answer 2

Guide:

let $a = \lim_{n \to \infty} a_n$,

$\forall \epsilon>0$, we can find $N>0$, such that $n > N$, $|a_n-a| < \epsilon$.

Now, given $\epsilon$, think of how to pick $M>0$, such that

$$n > M, |a_{n+1}-a| < \epsilon$$

Try to find such $M$ in terms of $N$.

Remark: A convergence sequence need not be monotone.

Answer 3

Let $L$ such that $\lim_{n\rightarrow\infty}a_n=L$. Let $\epsilon>0$ so, exists a natural number $N$ such that $$ n\geq N\Rightarrow|a_n-L|<\epsilon $$ Then $n+1>n\geq N$, so $|a_{n+1}-L|<\epsilon$.

Answer 4

Different method which uses somewhat more machinery: consider $a_{n+1} - a_n$, which tends to $0$ because $(a_n)$ is convergent and hence Cauchy. Since $(a_n)$ is convergent, the sequence $(a_{n+1})$ must therefore be convergent and must tend to the same limit, by the following theorem which you should make sure you can prove:

If $b_n - a_n \to c$, and $(a_n)$ converges to $a$, then $(b_n)$ is convergent and converges to $c-a$.

Answer 5

In the classical $\varepsilon >0$ and triangle inequality form, if $\lim\limits_{n\rightarrow\infty}a_n=a$: $$|a_{n+1} - a|=|a_{n+1} - a_n +a_n- a|\leq |a_{n+1} - a_n| +|a_n- a|\leq \\|a_{n+1} - a_n| +\frac{\varepsilon}{2} \leq ...$$ because a converging sequence is also a Cauchy sequence $$...\leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2}=\varepsilon$$ Similarly, it can be proved that $\lim\limits_{n\rightarrow\infty}a_{n+k}=a, \forall k\geq 0$.