Notation of sequence and limit convergence

Question

Trying to show that if the sequence $\{a_n\}$ converges to $L$ then $\lim_{n\rightarrow \infty} a_n = \lim_{n\rightarrow \infty} a_{n+1} = L$

Here is a proof attempt:

Since $\{a_n\}$ converges to $L$ we have that $\lim_{n\rightarrow \infty} a_n = L$

I use the definition of the limit at infinity:

Given any $\epsilon >0$, there exists integer $N$ such that if $n>N$ then $|a_n - L|<\epsilon\quad $ (Equation 1)

It holds that $n+1>n$ thus:

Given any $\epsilon >0$, there exists integer $N'$ (equal to $N$) such that if $n+1>N'$ then $|a_{n+1} - L|<\epsilon \quad $ (Equation 2)

Thus $\lim_{n\rightarrow \infty} a_{n+1} = L \quad $ [Q.E.D]

Issue No.1. My first point of confusion is Equation 2. Obviously $n+1>n$ but how do I justify precisely that $|a_{n} - L|<\epsilon$ becomes $|a_{n+1} - L|<\epsilon$ ?

Issue No.2. It holds that $n+m > n$ for any $m>0$. Thus we get:

Given any $\epsilon >0$, there exists integer $N''$ (equal to $N$) such that if $n+m>N''$ then $|a_{n+m} - L|<\epsilon \quad $ (Equation 3)

Can I conclude that $\lim_{n\rightarrow \infty} a_{n+m} = L$ for any $m>0$ ?

Issue No.3. It seems to me that if $\{a_n\}$ converges to $L$ then $\lim_{n\rightarrow \infty} a_{n+m} = L$ for any $m \in \mathbb{Z}$. Is that the case? If so, how should I go about proving this?

Answer 1

Your proof is essentially correct. To make it clearer you should keep the same $\epsilon$ throughout the proof. You want to show that given any $\epsilon > 0$, you can find $N$ such that $n > N \implies |a_{n+1}-L| < \epsilon$ :

Take $\epsilon > 0$. Since $\{a_n\} \to L$, we know that we can find $N_{\epsilon} \in \mathbb{N}$ such that for any $n > N_{\epsilon}$ we have $|a_n - L| < \epsilon$. Thus, for any $n$ such that $n+1>N_{\epsilon}$ we also have $|a_{n+1}-L|<\epsilon$, meaning that $\{a_{n+1}\} \to L$. $\square$

You can easily adapt this proof to answer your other issues. What might make this proof confusing is how obvious it seems. A better approach would be to use a more general result, such as the one suggested by @DavideMasi. You can prove such a result without much more work ; see for example this question.

Answer 2

Issue no 1:

Technically $|a_n -L|< \epsilon$ does not "become" $|a_{n+1} - L |< \epsilon$

You have that there exists any $N$ so that for any $n> N$ you have $|a_n-L| < \epsilon$. But this holds for any $n> N$ so if you have $n, m , h, j, \mu, \omicron, 10^n, \text{foo} > N$ then you have $|a_n-L| < \epsilon,|a_m-L| < \epsilon,|a_h-L| < \epsilon,|a_j-L| < \epsilon,|a_{\mu}-L| < \epsilon,|a_{\omicron}-L| < \epsilon,|a_{10^n}-L| < \epsilon,|a_{\text{foo}}-L| < \epsilon$. These are an infinite class of "different" statements but they are all true.

Issues No. 2 and 3:

Yes to both.

Furthermore... in the comments Davide Masi refers to a fact that if a sequence converges any subsequence will converge to the same limit. So if the sequence $a_1, a_2, a_3, ......., a_n, .......$ converges to $L$ then the subsequence $a_2, a_3, a_4,......., a_{n+1},.... $ will converge to $L$.

We can prove this fact (and we should NEVER use facts that haven't been proven) by pretty much the same argument.

If $a_{m_i}$ is a subsequence of $a_i$ that means each $m_i\in \mathbb N$ and $m_i < m_j \iff i < j$

(For example if if $a_i= \frac 1n$ so $\{a_i\}= \frac 11, \frac 12, \frac 13, \frac 14, etc.$ and if we take the subsequence $\frac 11, \frac 13, \frac 15= \{a_1, a_3, a_5,...\} = \{a_{2i-1}\}$ so $m_i = 2i -1$)

(In general $a_{m_i} = a_{m_1}, a_{m_2}, a_{m_3}... = a_2, a_5, a_6, a_8, a_{12}...$ where $m_1,m_2, m_3, m_4, m_5... = 2,5,6,8,12,....$ for some "subpattern")

Therefore for any $N$ where $1,2,3,4,....., N-1 < N$. $N=1$, and $N+1, N+2, .... > N$. there is a $M$ where $m_1, m_2, m_3, ....m_{M-1} < N$ and $N \le m_M$.

(For example if we have $a_i= \frac 1i$ and so $\{a_n\} = \{\frac 11,\frac 12,\frac 13, \frac 14, ...\}$ and $a_{m_i} = a_{i^2}$ so $\{a_{m_i}\}= \{\frac 11, \frac 14, \frac 19, \frac 1{16}$. And if $N = 67$ then so for $n< N$ we have $a_n \in \frac 11, \frac 12, ...., \frac 1{65}, \frac 1{66}$ and for $n\ge N$ we have $a_n = \frac 1{67}, \frac1{68}, \frac 1{69}$, the we can have $M =9$ so that $a_{m_1}=a_1=\frac 11, a_{m_2}=a_4 =\frac 14,.......a_{m_8}=a_{64}=\frac 1{64}$ where $i < 9=M$ and $m_i < 64 = N$.)

So if $a_n$ is a sequence that converges to $L$ and $a_{m_n}$ is a subsequence then:

For any $\epsilon > 0$ there is an $N$ so that for all $n> N$ we have $|a_n - L | < \epsilon$. So if $M$ is such that $m_M \ge N$ then for all $n > M$ then $m_n > N$ so $|a_{m_n} - L| < \epsilon$.

So $a_{m_i}$ converges to $L$.