Question

Prove \(lim_{n-->\infty} \dfrac{1}{n+1} =0\)
Please help

Answer 1

We know that
\[lim_{n\rightarrow \infty}\dfrac{1}{n}=0\]
but how to link them together?

Answer 2

Make a comparison:
\[\frac{1}{n+1}\le\frac{1}{n}\]
If you can prove the limit for \(\dfrac{1}{n}\), you can prove the the other limit.

Answer 3

I am allowed to use the fact that lim (1/n) =0

Answer 4

Can I use squeeze theorem? but if using it, I have to prove that lim (1/n+1) exists first.

Answer 5

What kind of definition are you using? The \(\epsilon-N\) for sequences or \(\epsilon-\delta\) for functions?

Answer 6

I am sorry, I don't get you.

Answer 7

You are talking about epsilon?
I have to use limit theorem to prove it

Answer 8

I can't ( not allowed) use definition of limit.

Answer 9

The \(\epsilon-N\) definition of the limit applies to sequences. It says
\[\lim_{n\to\infty}a_n=L\]
means there exists \(N\) such that \(N\ge n\) implies \(|a_n-L|<\epsilon\) for any \(\epsilon>0\).

In this case, you would want to find such an \(N\) that makes \(\left|\dfrac{1}{n+1}\right|<\epsilon\). You can show this easily using the fact that \(n+1>n\) for all \(n\), which gives
\[\frac{1}{n+1}<\frac{1}{n}<\epsilon~~\implies~~N=\left\lceil\frac{1}{\epsilon}\right\rceil\]
where \(\lceil x\rceil\) denotes the ceiling function, which gives the smallest integer greater than \(x\).

Answer 10

I got you. Thank you :)

Answer 11

You're welcome!

Answer 12

You can also do this, right?
Proof: \[\text{ Let } \epsilon>0 \text{. Set } \\ M=max\left\{ -1, \frac{1-\epsilon}{\epsilon} \right\} \\ \text{ If } n>M \text{, then } n>-1 \text{ and } n> \frac{1- \epsilon}{\epsilon} \text{, } \\ \\ \text{ so } n > \frac{1- \epsilon}{\epsilon} \\ \epsilon n > 1- \epsilon \\ \epsilon n +\epsilon > 1 \\ \epsilon(n+1)>1 \\ \epsilon>\frac{1}{n+1 } \\ |\frac{1}{n+1}-0| < \epsilon \]