prove that $\lim_{n\to +\infty }a_n=a=\lim_{n\to+\infty }a_{n+k}$, $k\ge 1$ Let $\lim_{n\to +\infty}a_n=a$.
In the $\lim_{n\to +\infty }a_n=a$,
I set where $n$, $n+k$ then I get $$\lim_{n+k\to +\infty }a_{n+k}=a\iff\lim_{n\to+\infty -k}a_{n+k}=a\iff\lim_{n\to +\infty }a_{n+k}=a.$$
Am I right ??
 A: We do not necessarily define expressions like $+\infty-k$. Instead, I prefer to consider $$\tag1\lim_{\square\to+\infty}\square$$  a single opaque symbol where the two places symbolized by $\square$ can be filled suitably, namely the lower one by a variable name and the other one by an expression depending on that variable. This looks like, but differs from
$$\tag2 \lim_{\square\to\square}\square$$
(where the additional square can be filled with an expression for a real number - infinity is not a real number!) in an essential way:
We have
$$ \lim_{x\to a}f(x)=b\iff \forall\epsilon>0\colon\exists\delta>0\colon\forall y\colon|y-a|<\delta\to|f(x)-b|<\epsilon $$
but instead of 
$$ \lim_{x\to +\infty}f(x)=b\iff \forall\epsilon>0\colon\exists\delta>0\colon\forall y\colon|y-\infty|<\delta\to|f(x)-b|<\epsilon $$
we have
$$ \tag3\lim_{x\to +\infty}f(x)=b\iff \forall\epsilon>0\colon\exists M\colon\forall y\colon y>M\to|f(x)-b|<\epsilon.$$
One can support the view that $(1)$ is a special case of $(2)$, but that requires viewing the structure of $\mathbb R$ (together with $\infty$) as a slightly different space: not endowed with the usual metric and not a field or even an additive group! Before you are sure what that means and how to formalize the treatment of expressions such as $\infty-k$ suitably, I recommend to always keep the  difference between $(1)$ and $(2)$ in mind!
That being said, I suggest to prove the original claim directly from $(3)$.
