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Prove that if $$ \lim\limits_{n \rightarrow \infty} \frac{S_n-s}{S_n+s} = 0$$ then $$\lim\limits_{n \rightarrow \infty} S_n = s$$

Hint: Define $t_n = \frac{S_n -s}{S_n + s}$ and solve for $S_n$

By the hint: $$t_n = \frac{S_n -s}{S_n + s}$$ $$(S_n + s)t_n = S_n -s$$ $$S_n(t_n-1)= - s -st_n$$ $$S_n= -s \cdot \frac{1+t_n}{t_n-1}$$

$$\lim\limits_{n \rightarrow \infty} S_n= -s \cdot \frac{1+\lim\limits_{n \rightarrow \infty} t_n}{\lim\limits_{n \rightarrow \infty} t_n-1}$$

As $ \lim\limits_{n \rightarrow \infty} \frac{S_n-s}{S_n+s} = \lim\limits_{n \rightarrow \infty} t_n = 0$, it follows:

$$\lim\limits_{n \rightarrow \infty} S_n= s$$

Is my argumentation correct/appropriate?Anything needs to be added?

Much appreciated for your input.

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    $\begingroup$ You need to be very careful, e.g. when dividing by $t_n-1$ that you are not dividing by zero. So you need to take $n$ large enough that $t_n$ is bounded away from $1$ - which is possible because it has limit equal to $0$. $\endgroup$ – Mark Bennet Aug 6 '17 at 11:18
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    $\begingroup$ @rei I think your proof is right and nice! Since $t_n\rightarrow0$, we have no problems with denominator. $\endgroup$ – Michael Rozenberg Aug 6 '17 at 11:18
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I think your proof is right and nice! Since $t_n\rightarrow0$, we have no problems with denominator.

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Alternatively: Let $S_n=T_n-s$, then: $$\lim_\limits{n\to\infty} \left(1-\frac{2s}{T_n}\right)=0 \Rightarrow \lim_\limits{n\to\infty} T_n=2s \Rightarrow $$ $$\lim_\limits{n\to\infty} (S_n+s)=2s \Rightarrow \lim_\limits{n\to\infty} S_n=s.$$

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To fill in the small gap pointed out by @Mark Bennet in the comment, one might use the following argument:

Since $\lim_{n\to \infty} t_n=0, \forall\varepsilon\gt 0, \exists N\in \Bbb N$ such that $\forall k\ge N, |t_k-0|=|t_k|\lt \varepsilon$. Now, if one chooses $\varepsilon=1$, then from some $m\in \Bbb N$ onwards, $|t_n|\lt 1$ and therefore $|t_n-1|$ is non-zero.

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