## anonymous 5 years ago how can we prove that a(n) = n/n+1 converges by using the definition? I know how to find the limit, yet I can't seem to find a way to prove it using the definition

1. anonymous

I might be able to help you. I would consider the following:$\lim_{n->\infty}\frac{n}{n+1}=\lim_{n->\infty}\frac{1}{1+\frac{1}{n}}$and from limit laws, $\frac{1}{1+\lim_{n \rightarrow \infty} \frac{1}{n}}$Now, for $\lim_{n \rightarrow \infty}\frac{1}{n}$choose $\epsilon >0$ small. Let N be the first integer greater than $\frac{1}{\epsilon}$Then$\forall n>N \rightarrow n>\frac{1}{\epsilon} \iff \forall n>N \rightarrow \epsilon >\frac{1}{n} \iff \frac{1}{n}< \epsilon$Since n>0,$0<\frac{1}{n}< \epsilon \iff 0<\frac{1}{n} - 0 < \epsilon$ and so, by definition of the limit,$\frac{1}{n} \rightarrow 0$ as $n \rightarrow \infty$

2. anonymous

So,$\lim_{n->\infty}\frac{n}{1+n}=\lim_{n->\infty}\frac{1}{1+\frac{1}{n}}=\frac{1}{1+\lim_{n->\infty}\frac{1}{n}}=\frac{1}{1+0}=1$

3. nowhereman

You even do not need that limit law, if you already know the limit. Just calculate $|1-\frac{n}{1+n}| = \frac{1}{1+n}$ so need only find n for ε so that $\frac 1 {1+n} < ε ⇔ n > \frac 1 ε - 1$ Thus if you choose $n = \frac 1 ε$ everythings fine.

4. anonymous

thank you :)