Prove that $\lim_{n\to\infty}b_n = a$ Let $\lim_{n\to\infty}a_n = a$. Let $(b_n)$ be a sequence satisfying $b_{n+k} = a_{n+l}$ for some $k,l\in\mathbb N$ and all $n\in\mathbb N$. Prove that $\lim_{n\to\infty}b_n = a$
...Well, first, I want to understand the statement exactly. Does it mean that if we have an arithmetic sequence, which is equal to another arithmetic sequence, they have the same limit? - If I notice that both sequences are in $\mathbb N$, am I right?
Well, after understood the statement, I should prove it... something which I have never done in my life and I should learn.
 A: The relation between the sequences $(b_n)$ and $(a_n)$ is that they look the same after some index $N \in \mathbb{N}$. As an example consider the sequence defined by $x_n = n$ and let $y_{n+2} = x_{n+3}$. The sequence $y$ would be 
$$ y_3 = x_6 = 6, \ y_4 = x_7 = 7$$
and so on. The shifting of the indices only affects the initial part of the sequences, which has no effect on the convergence of the sequence.
However there are some technical nuances when thinking about how to prove this "trivial" fact. By convergence of the sequence $(a_n)$ we have
$$ \forall \varepsilon > 0 : \exists N \in \mathbb{N} : n \geq N : |a_n - a| < \varepsilon$$
Well if we directly plug in the relation between $a_n$ and $b_n$ then we would have
$$ |a_{n + (l - l)} - a| = |b_{n - l + k} - a|$$
Now if we set $b_{n - l + k} = b_m$ we may note that the index relation above will be given as 
$$n \geq N \iff m + l - k \geq N \iff m \geq N - l + k$$
So the proof has written itself! Let $\varepsilon > 0$ choose $M = \max\{ N(\varepsilon) - l + k, 1 \}$ by the above
$$ \forall \varepsilon > 0 : \exists M \in \mathbb{N} : m \geq M : |b_m - a|  < \varepsilon$$
So
$$ \lim_{m \to \infty} b_m = a \iff \lim_{n \to \infty} b_{n + l - k} = a$$
A: If $l\geq k$ then $b_{n}=b_{(n-k)+k} = a_{(n-k)+l}=a_{n+l-k}$ thus $b_n$ is a subsequence of $a_n$ so $b_n\to a$
Likewise if $l < k$  then for $n>k-l$, we have $b_n$ is a subsequence of $a_n$.
A: By the definition of the limit of a sequence, we have that for every $\varepsilon>0$ there exists $N(\varepsilon)\in\mathbb N$ such that
$$
|a_n-a|<\varepsilon
$$
if $n\ge N(\varepsilon)$.
We have that
$$
|b_{n+k}-a|=|a_{n+l}-a|<\varepsilon
$$
as soon as $n\ge N(\varepsilon)-l$. Hence, for every $\varepsilon>0$ there exists $M(\varepsilon)$ such that
$$
|b_n-a|<\varepsilon
$$
as soon as $n\ge M(\varepsilon)$. That is what we wanted to prove. We have that $\lim_{n\to\infty}b_n=a$.
A: According to the given statement, nothing is said about arithmetic sequence (recall the definition of an arithmetic sequence). 
Let $a_{n} \to a$ as $n \to \infty$; let $k, l \in \mathbb{N}$; let $b_{n+k} := a_{n+l}$ for all $n \geq 1$; then $|b_{n+k}-a| = |a_{n+l}-a|$ for all $n \geq 1$. If $\varepsilon > 0$, then there is some $N \geq 1$ such that $|a_{n}-a| < \varepsilon$ for all $n \geq N$, implying that for all $n \geq \max \{ N, l+1 \}$ we have $|a_{n+l} - a| = |b_{n+k} - a| < \varepsilon$.
