My method of proving divergence

**sfspitfire23** · March 12th 2010, 05:17 PM

Prove: $\frac{n^2+1}{n-2}$ diverges

For every L there exists an epsilon greater than zero, for every N greater than 0, n>0.

So, $|\frac{n^2+1}{n-2}-L|>\epsilon$

Pf: Let $L$ be given. Set $\epsilon=.001$ and let $N>0$ be given. There exists $n_0>N$ s.t. $n_0=3$ . Then, if $|\frac{n_0^2+1}{n_0-2}-L|<\epsilon$ , we would have $|10-L|<\epsilon$ $\implies$ $-.001<L-10<.001 \implies 9.999<L<10.001$ . But then if $|\frac{n_{0+1}^2+1}{n_{0+1}-2}-L|<\epsilon$ we would have $|8.5-L|<\epsilon$ . So, $8.499<L<8.501$ . But then $L$ would be in different intervals. Thus the sequence diverges as desired.

Is this the correct way to go about this?

Thx!

**Drexel28** · March 12th 2010, 05:21 PM

Originally Posted by sfspitfire23

Prove: $\frac{n^2+1}{n-2}$ diverges

For every L there exists an epsilon greater than zero, for every N greater than 0, n>0.

So, $|\frac{n^2+1}{n-2}-L|>\epsilon$

Pf: Let $L$ be given. Set $\epsilon=.001$ and let $N>0$ be given. There exists $n_0>N$ s.t. $n_0=3$ . Then, if $|\frac{n_0^2+1}{n_0-2}-L|<\epsilon$ , we would have $|10-L|<\epsilon$ $\implies$ $-.001<L-10<.001 \implies 9.999<L<10.001$ . But then if $|\frac{n_{0+1}^2+1}{n_{0+1}-2}-L|<\epsilon$ we would have |8.5-L|<\epsilon. So, $8.499<L<8.501$ . But then $L$ would be in different intervals. Thus the sequence diverges as desired.

Is this the correct way to go about this?

Thx!

It looks as though you have proved that the sequence does not converge to any $L\in\mathbb{R}$ . Is that what they want you to do?

More informative would be to prove the rigorous idea that $\frac{n^2+1}{n-2}\to\infty$ .

To do this first note that $\frac{n^2+1}{n-2}\geqslant n+\frac{1}{n}\geqslant n$ and so given any $T>0$ we may choose $N\in\mathbb{N}$ (namely $N=\left\lceil T\right\rceil$ ) such that $N\leqslant n\implies T<n$ . The conclusion follows.

**Tinyboss** · March 12th 2010, 05:21 PM

It doesn't need to be that complicated. Try replacing the numerator with something smaller, and the denominator with something larger. Choose your replacements in such a way that the quotient simplifies to something that obviously diverges. Then you'll know that the original sequence is larger than a sequence that diverges.

**sfspitfire23** · March 12th 2010, 05:31 PM

Ah ok....

How would you recognize that
$ \frac{n^2+1}{n-2}\geqslant n+\frac{1}{n}\geqslant n $ ?

Experience? haha

**Drexel28** · March 12th 2010, 05:32 PM

Originally Posted by sfspitfire23

Ah ok....

How would you recognize that
$ \frac{n^2+1}{n-2}\geqslant n+\frac{1}{n}\geqslant n $ ?

Experience? haha

That should be pretty apparent. Making the denominator smaller makes the whole fraction larger so increasing the denominator makes the fraction smaller, etc.

**sfspitfire23** · March 12th 2010, 05:40 PM

So, to show divergence to infinity I can find a smaller sequence than my sequence and play with that.

To show convergence I can show a larger sequence than my sequence converges

Yeah?

**Defunkt** · March 13th 2010, 10:19 AM

Originally Posted by sfspitfire23

So, to show divergence to infinity I can find a smaller sequence than my sequence and play with that.

To show convergence I can show a larger sequence than my sequence converges

Yeah?

The first is correct - if your sequence is greater than a sequence which "goes" to infinity, then so does your sequence.

The second conclusion is not always correct, though. Consider $a_n = 1+\frac{1}{n}, ~ b_n = sin(n)$ , then obviously $b_n < a_n ~ \forall n \in \mathbb{N}$ and $a_n \to 1$ but $b_n$ does not converge.

It is correct, though, that a bounded, monotone sequence converges.

Thread: My method of proving divergence

LinkBack

Thread Tools

Display

My method of proving divergence

Similar Math Help Forum Discussions

Method of difference (proving)

Trouble proving convergence/divergence

Element Method for proving...

Proving convergence of divergence of a series.

Newton method : divergence

Search Tags