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Suppose that $|S_n - S| \leq t_n$ for large $n$ and $\lim\limits_{n \rightarrow \infty} t_n =0$. Show that $\lim\limits_{n \rightarrow \infty} S_n =S$.


As the distance between $S_n$ and the finite number $S$ is bounded above by $t_n$. When $n \rightarrow \infty$ and $t_n$ converges to $0$, the distance $|S_n - S|$ has to converge to zero.

$$\lim\limits_{n \rightarrow \infty} |S_n - S|=0 $$

As the distance is null

$$S_n - S=0 $$

As $S$ is finite, it follows that, when $n$ is large, $S_n =S$


Question:
Is my argumentation appropriate/correct? How would you show this?

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    $\begingroup$ No, it isn't correct. $S_n\to S$ doesn't mean $S_n=S$ for $n$ large. Note that $1/n\to 0$ but $1/n\ne 0,\forall n\in\mathbb{N}.$ $\endgroup$ – mfl Aug 4 '17 at 21:14
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Your argument might fail because it implicitly assumes that there exists $N$ such that $S_n=S$ for $n\geq N$. For a counter example, we could take $S_n = 1/n^2$, and $t_n = 1/n$, then both $(S_n)$ and $(t_n)$ converge towards $0$ and $|S_n| \leq t_n$ for every $n$, however $S_n \neq 0$ for every $n$.

So, let us assume that there exists $M>0$ such that $|S_n-S|\leq t_n$ for every $n\geq M$ and $\lim_{n\to \infty}t_n =0$.

Proof of $\lim_{n\to\infty}S_n=S$:
Let $\epsilon >0$, then there exists $N$ such that $t_n<\epsilon$ for all $n\geq N$, it follows that for every $n\geq \max\{N,M\}$ we have $|S_n-S|\leq t_n <\epsilon$. Since this is true for every $\epsilon>0$, it follows that $\lim_{n\to \infty} S_n = S$.

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  • $\begingroup$ thx for the input. Could you elaborate on "we prove that $\lim_{n \rightarrow \infty} S_n = 0$"? $\endgroup$ – rei Aug 5 '17 at 1:31
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    $\begingroup$ @rei sorry it was a typo. $\endgroup$ – Surb Aug 5 '17 at 12:41
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Your argument is not quite correct because we could have the limit of $(S_n)$ is $S$ without having exactly $S_n = S$. But $\lim\limits_{n \rightarrow \infty} |S_n - S|=0$ is enough to prove the claim. Use the definition of limit

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