If the sequence $ x_{n} $ converges to L, then $\lim_{k\to \infty}x_{k+1} = L $ Can someone read this proof and let me know if it is correct?
If the sequence $ x_{n} $ converges to $L$, then $$\lim_{k\to \infty}x_{k+1} = L $$  
Proof.
Let $ \epsilon  > 0$, and suppose $\lim_{k\to \infty}x_{k} = L$.
Then by supposition, there is some $M' \in N$ such that $\lvert x_{k'}-L\rvert < \epsilon$ for $k' \geq M'$.
Let $M := M'-1$. 
Then if $k \geq M$ we have $k+1 \geq M'$ so we can apply $k' = k+1$ to $\lvert x_{k'}-L\rvert < \epsilon$ for $k' \geq M'$
Hence we have $\lvert x_{k+1}-L\rvert < \epsilon$ for $k+1 ≥ M'$.
 A: It's almost correct. You are making it more complicated than it actually is. It should be obvious that
$$\forall\varepsilon{>}0\,\exists n{>}0\,\forall k{\ge}n:|x_k-L|<\varepsilon\implies\forall\varepsilon{>}0\,\exists n{>}0\,\forall k{\ge}n:|x_{k+1}-L|<\varepsilon$$
simply because $k+1\ge k\ge n$. This corresponds to $M=M'$ in your proof, but in your case $M'$ could be $-1$ if $M=0$ which is not a natural number (or $0$ if $M=1$ in case you don't consider $0$ a natural number).
A: The proof seems fine, but maybe a more general version could be useful in different contexts.
Suppose $n\mapsto f(n)$ is a strictly increasing map $f\colon \mathbb{N}\to\mathbb{N}$. If $(x_n)_{n\in\mathbb{N}}$ is a sequence, then, setting $y_n=x_{f(n)}$, the sequence $(y_n)_{n\in\mathbb{N}}$ is called a subsequence.
Note that, for $f$ strictly increasing, we must have $f(n)\ge n$, for every $n$. Indeed, $f(0)\ge0$, which is obvious. If $f(n)\ge n$, then $f(n+1)>f(n)\ge n$, so $f(n+1)\ge n+1$.
Theorem. If $(x_n)_{n\in\mathbb{N}}$ is a sequence that converges to $L$, then every subsequence converges to $L$.
Proof. Suppose the subsequence is defined by the strictly increasing function $f$ by $y_n=x_{f(n)}$. Fix $\varepsilon>0$. Then there exists $M\in\mathbb{N}$ such that, for $n>M$, $|x_n-L|<\varepsilon$. Then, if $n>M$, we have
$$
|y_n-L|=|x_{f(n)}-L|<\varepsilon
$$
because $n>M$ implies $f(n)>f(M)\ge M$. QED
Yours is the simple case in which $f(n)=n+1$.
