1. Limits of sequences

Describe the limiting behaviour of the following sequences:

1) $\displaystyle \dfrac{\ln n}{n^a}$, $\displaystyle a>0$

I think its 0. I worked this out because I know $\displaystyle n^a$ grows faster than $\displaystyle \ln n$. Is that good enough to do that in a exam situation, or would you need to prove it.

2) $\displaystyle \dfrac{n\sin (n\pi/4)}{\sqrt{n^2+1}}$

I think this diverges, but I'm not too sure how to show it.

3) $\displaystyle \dfrac{n+\cos n\pi}{n-\cos n\pi}$

Would I divide top and bottom by $\displaystyle n$ and so the sequence would converge to -1

2. When you say "the limiting behaviour", I assume you mean the limit as $\displaystyle n \to \infty$.

$\displaystyle \lim_{n \to \infty}\frac{\ln{n}}{n^a}$ tends to $\displaystyle \frac{\infty}{\infty}$, so you can use L'Hospiatal's Rule.

$\displaystyle \lim_{n \to \infty}\frac{\ln{n}}{n^a} = \lim_{n \to \infty}\frac{\frac{1}{n}}{an^{a - 1}}$

$\displaystyle = \lim_{n \to \infty}\frac{1}{an^a}$

$\displaystyle = 0$.

2. Note that $\displaystyle \sin{x} \leq 1$.

Therefore $\displaystyle \frac{n\sin{\left(\frac{n\pi}{4}\right)}}{\sqrt{n^ 2 + 1}} \leq \frac{n}{\sqrt{n^2 + 1}}$.

So if the second sequence is convergent, so is the first.

$\displaystyle \lim_{n \to \infty}\frac{n}{\sqrt{n^2 + 1}} = \lim_{n \to \infty}\frac{1}{\sqrt{1 + \frac{1}{n^2}}}$

$\displaystyle = 1$.

Since $\displaystyle \frac{n}{\sqrt{n^2 + 1}}$ is convergent, so is $\displaystyle \frac{n\sin{\left(\frac{n\pi}{4}\right)}}{\sqrt{n^ 2 + 1}}$.

3. $\displaystyle \lim_{n \to \infty}\frac{n + \cos{(n\pi)}}{n - \cos{(n\pi)}} =\lim_{n \to \infty} \frac{n - \cos{(n\pi)} + 2\cos{(n\pi)}}{n - \cos{(n\pi)}}$

$\displaystyle = \lim_{n \to \infty}\left(1 + \frac{2\cos{(n\pi)}}{n - \cos{(n\pi)}}\right)$

$\displaystyle = 1$ since in the fraction the top oscillates while the bottom grows without bound.

3. Originally Posted by Prove It
2. Note that $\displaystyle \sin{x} \leq 1$.

Therefore $\displaystyle \frac{n\sin{\left(\frac{n\pi}{4}\right)}}{\sqrt{n^ 2 + 1}} \leq \frac{n}{\sqrt{n^2 + 1}}$.

So if the second sequence is convergent, so is the first.

$\displaystyle \lim_{n \to \infty}\frac{n}{\sqrt{n^2 + 1}} = \lim_{n \to \infty}\frac{1}{\sqrt{1 + \frac{1}{n^2}}}$

$\displaystyle = 1$.

Since $\displaystyle \frac{n}{\sqrt{n^2 + 1}}$ is convergent, so is $\displaystyle \frac{n\sin{\left(\frac{n\pi}{4}\right)}}{\sqrt{n^ 2 + 1}}$.
Wrong. It does not converge since $\displaystyle \sin(n\pi/4)$ never settles down this cycles through values close to $\displaystyle 0, \sqrt{2}, 1, \sqrt{2}, 0,-\sqrt{2},-1,-\sqrt{2}$.

It's a sequence not a series.

CB

4. Originally Posted by CaptainBlack
Wrong. It does not converge since $\displaystyle \sin(n\pi/4)$ never settles down this cycles through values close to $\displaystyle 0, \sqrt{2}, 1, \sqrt{2}, 0,-\sqrt{2},-1,-\sqrt{2}$.

It's a sequence not a series.

CB
Yes, you are right, sorry. All I did was show that the values are always no greater than 1.

5. Originally Posted by Prove It
3. $\displaystyle \lim_{n \to \infty}\frac{n + \cos{(n\pi)}}{n - \cos{(n\pi)}} =\lim_{n \to \infty} \frac{n - \cos{(n\pi)} + 2\cos{(n\pi)}}{n - \cos{(n\pi)}}$

$\displaystyle = \lim_{n \to \infty}\left(1 + \frac{2\cos{(n\pi)}}{n - \cos{(n\pi)}}\right)$

$\displaystyle = 1$ since in the fraction the top oscillates while the bottom grows without bound.
Simpler (and closer to what the OP tried to do):

$\displaystyle \dfrac{n+\cos(n\pi)}{n-\cos(n\pi)}=\dfrac{1+\frac{\cos(n\pi)}{n}}{1-\frac{\cos(n\pi)}{n}}$

So as top and bottom of the right hand side both go to $\displaystyle 1$ as $\displaystyle n\to \infty$ the limit of the whole thing is $\displaystyle 1$.

CB