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 $n \geq N$ $$n > 1/\epsilon - 1 \geq N $$


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

thx for the help.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.