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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!

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    $\begingroup$ You need to use the definition of convergence. In this definition there is a statement like $\exists n \in \mathcal N$ such that $\forall k>n$. If you have this n for $(a_{n+1})$ you can easily find a "n'" that will fulfill the condition for $(a_n)$. $\endgroup$ – Max Ft Mar 27 '18 at 19:22
  • $\begingroup$ @ Max Ft: I concur with that idea! $\endgroup$ – Omojola Micheal Mar 27 '18 at 19:29
  • $\begingroup$ @Mike Please remember that you can choose an answer among the given if the OP is solved, more details here meta.stackexchange.com/questions/5234/… $\endgroup$ – gimusi Mar 28 '18 at 20:06
  • $\begingroup$ @ gimusi: Thank you very much for that! Yes, your answer really helped me! Others also did! I appreciate you all! $\endgroup$ – Omojola Micheal Mar 30 '18 at 19:41
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Note that

  • $|a_n-L|<\epsilon$ for $n>N\implies |a_{n+1}-L|<\epsilon$ for $n+1>N$
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  • $\begingroup$ @ gimusi: Thank you very much! I respect you in particular! My question is: how are we sure that $a_{n}$ and $a_{n+1}$ converge to the same point? I am aware that we can show it with $a_{n}=\left( 1+\frac{1}{n}\right)^{n}$ and $a_{n+1}=\left( 1+\frac{1}{n}\right)^{n+1}$ also. $\endgroup$ – Omojola Micheal Mar 27 '18 at 19:25
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    $\begingroup$ @Mike Hi Mike, simply by the definition note that, by hypotesis, $\forall \epsilon >0$ we can find $N$ such that $\forall n>N$ such that $|a_n-L|<\epsilon$ then $\forall n+1>N \implies |a_{n+1}-L|<\epsilon$ therefore also $a_{n+1}\to L$. Intuitively think to the fact that $a_{n+1}$ is just $a_n$ shifted of 1 thus the limit is the same. Your example should be $a_{n+1}=\left( 1+\frac{1}{n+1}\right)^{n+1}$. $\endgroup$ – gimusi Mar 27 '18 at 19:30
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    $\begingroup$ @ gimusi: Oh! I get it! Thanks for that correction! $\endgroup$ – Omojola Micheal Mar 27 '18 at 19:33
  • $\begingroup$ @Mike You are welcome! Well done, Bye $\endgroup$ – gimusi Mar 27 '18 at 19:36
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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 $|a_{n+1}-L|<\epsilon$.

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  • $\begingroup$ @ Gödel: This is very nice! I love it! Thanks a lot! $\endgroup$ – Omojola Micheal Mar 27 '18 at 19:28
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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 $c-a$.

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  • $\begingroup$ @ Patrick Stevens: This solution is unique! Thank you! $\endgroup$ – Omojola Micheal Mar 27 '18 at 21:21
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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.

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  • $\begingroup$ @ Siong Thye Goh: You are good! Thank you very much! $\endgroup$ – Omojola Micheal Mar 27 '18 at 19:27
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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 $\lim\limits_{n\rightarrow\infty}a_{n+k}=a, \forall k\geq 0$.

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  • $\begingroup$ @ rtybase: I admire your solution! Regards! $\endgroup$ – Omojola Micheal Mar 27 '18 at 21:20

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