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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 :(

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    When you don't know how to start, start with the definition. What does it mean to say that $b_n\to b$?2012-10-01

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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 $b_{n+M}\gt B$.

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    Manipulating formulas should often come more or less near the end,and you have figured out what's really going on. What's going on is that after a while, $b_n$ is small, so of course $b_{n+M}$ is small. Then filling in the $\epsilon$ stuff is easy.2012-10-01