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Find $\lim\limits_{n \rightarrow \infty} S_n$. Justify using definition of convergence/divergence of a sequence: $$S_n =2 + \frac{1}{n+1}$$

The limit: $$\lim\limits_{n \rightarrow \infty} S_n = \lim\limits_{n \rightarrow \infty} 2 + \frac{1}{n+1} = 2$$

Since $$|S_n -2| = |2 + \frac{1}{n+1} -2|= \frac{1}{1+n} < \epsilon$$

Hence if $\epsilon>0$, then $S_n$ holds with $s=2$ if $N\geq \frac{1}{\epsilon}$


The question I have regards the process to find the last part that I believe (hopefully) is correct: $N\geq \frac{1}{\epsilon}$

Given $$\frac{1}{1+n} < \epsilon$$ solving for n $$n > 1/\epsilon - 1 $$ Given that $n \geq N$ $$n > 1/\epsilon - 1 \geq N $$

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What is the process to find $N\geq \frac{1}{\epsilon}$ ?

thx for the help.

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  • $\begingroup$ How did you get that $\frac{1}{\varepsilon - 1} \geq N$? It's not always true that $a>b \land a \geq c \implies b \geq c$. $\endgroup$ – platty Aug 4 '17 at 22:57
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Let $N \geq \frac{1}{\varepsilon}$. Then we want to show that for any $n \geq N$, it is true that $|S_n - 2| \leq \varepsilon$. Indeed, $$|S_n - 2| = \left|2+\frac{1}{n+1} - 2\right| = \left|\frac{1}{n+1}\right|$$ But $n \geq N$, so $n+1 > N$ and $\frac{1}{n+1} < \frac{1}{N} = \frac{1}{\frac{1}{\varepsilon}} = \varepsilon$, as desired.

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