Question

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

Answer 1

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 \[n \rightarrow \infty\]

Answer 2

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

Answer 3

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.

Answer 4

thank you :)