Limit of a piecewise defined sequence

Question

If $S_n \to \ell$ as $n \to \infty$ prove that $t_n \to 3 \ell$ as $n \to \infty$ where $\displaystyle t_n = \begin{cases} n^2, & \mbox{if } n \le 1000 \\ 3S_n, & \mbox{if } n > 1000 \end{cases} $.

First of all, I can't construct an $\varepsilon$ that works. I thought I should say

$|n^2-3\ell|+|3S_n-3\ell| \le |3S_n-3\ell| = 3|S_n-\ell| < 3 \epsilon.$

$|3S_n-3\ell| -|n^2-3\ell| \le |3S_n-3\ell| = 3|S_n-\ell| < 3 \epsilon.$

Adding we get $|3S_n-3\ell|<\frac{3}{2}\epsilon$ and subtracting we get $|n^2-3\ell| <0$.

But I can't get the expression expression containing $n^2$ less than $\epsilon$.

Answer 1

You do not have to get terms of the form $n^2$ within $\epsilon$ of $3\ell$. The hard part of proofs like these is to find a number $N$ such that if $n\geq N$, then $|t_n-3\ell|<\epsilon$. If you find a value of $N$ that works, you can always make it bigger if you want. Therefore, you can just assume that $N> 1000$. Also since $S_n$ converges to $\ell$, there exists $N$ such that if $n\geq N$, then $|S_n-\ell|<\epsilon/3$.

Answer 2

It doesn't matter what $t_n$ does while $n<1000.$

$\forall \epsilon>0,\exists N>0$ such that $n>N \implies |t_n - \ell|<\epsilon.$