# Thread: My method of proving divergence

1. ## My method of proving divergence

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

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

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

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

Thx!

2. Originally Posted by sfspitfire23
Prove: $\displaystyle \frac{n^2+1}{n-2}$ diverges

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

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

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

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

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

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

3. 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.

4. Ah ok....

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

Experience? haha

5. Originally Posted by sfspitfire23
Ah ok....

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

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.

6. 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?

7. 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 $\displaystyle a_n = 1+\frac{1}{n}, ~ b_n = sin(n)$, then obviously $\displaystyle b_n < a_n ~ \forall n \in \mathbb{N}$ and $\displaystyle a_n \to 1$ but $\displaystyle b_n$ does not converge.

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