Prove: The tail end theorem:
For any positive integer $M$, if $b_n \to b$, then $b_{n+M} \to b$.
I do not know how to get started :(
Prove: The tail end theorem:
For any positive integer $M$, if $b_n \to b$, then $b_{n+M} \to b$.
I do not know how to get started :(
We have $\lim_{n\to\infty} b_n=b$ if and only if for every $\epsilon \gt 0$, there exists $N$ such that if $n\gt N$ then $|b_n-b|\lt \epsilon$.
Let $c_n=b_{n+M}$. Then if $n\gt N$, we have $n+M\gt N$, and therefore $|c_n-b|\lt \epsilon$. It follows that the sequence $(c_n)=(b_{n+M})$ has limit $b$.
Remark: The argument as written does not quite work if we want to allow the possibilities $b=\infty$ and $b=-\infty$. But it is not hard to write separate arguments along the same lines to take care of things if we want to allow infinite limits.
For example, for $+\infty$, we have $\lim_{n\to \infty} b_n =\infty$ iff for any $B$, there is an $N$ such that if $n\gt N$ then $b_n\gt B$. For the same $B$, if $n\gt N$ we will therefore have $b_{n+M}\gt B$.