r they convergent or not?

• May 16th 2008, 06:15 AM
szpengchao
r they convergent or not?
• May 16th 2008, 07:12 AM
flyingsquirrel
Hi

For the first one, you can try to show that $\displaystyle \left|\frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2}\right|\sim\frac{1}{n}$ for $\displaystyle n\to\infty$. For the second one, the ratio test may be helpful.

Not completely off-topic question : If $\displaystyle f \sim g$ do English speaking people say that $\displaystyle g$ is an approximation of $\displaystyle f$ ? Is there another way of saying this ?
• May 16th 2008, 07:38 AM
szpengchao
sorry
sorry, i found

the first one is less or equal to (n^3+8*n^2+1)/(n^4-n^2),
i think the later one is convergent. so the first one is convergent?
• May 16th 2008, 07:46 AM
Dr Zoidburg
Quote:

Originally Posted by flyingsquirrel
Not completely off-topic question : If $\displaystyle f \sim g$ do English speaking people say that $\displaystyle g$ is an approximation of $\displaystyle f$ ? Is there another way of saying this ?

The only other way I can think of, off-hand, is that $\displaystyle g$ approximates to $\displaystyle f$.
• May 16th 2008, 07:51 AM
flyingsquirrel
Quote:

Originally Posted by szpengchao
the first one is less or equal to (n^3+8*n^2+1)/(n^4-n^2),
i think the later one is convergent.

It's divergent. (and according to the hint I gave you, $\displaystyle \sum \frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2}$ can't be convergent) Anyway, the comparison test can also be used to show the divergence. :D
• May 16th 2008, 08:08 AM
szpengchao
ok
so how can u show that sequence tends to 1/n ????
• May 16th 2008, 08:16 AM
flyingsquirrel
Quote:

Originally Posted by szpengchao
so how can u show that sequence tends to 1/n ????

You can get it by showing that

$\displaystyle \frac{\left|\frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2}\right|}{\frac{1}{n}}=\left|\frac{n^4+ (-1)^n8n^3+n}{n^4+(-1)^{n+1}n^2}\right|$

tends to 1 as $\displaystyle n$ approaches $\displaystyle \infty$. (divide the numerator and the denominator by $\displaystyle n^4$)
• May 16th 2008, 12:00 PM
Mathstud28
Quote:

Originally Posted by szpengchao
so how can u show that sequence tends to 1/n ????

If $\displaystyle \lim_{x\to{c}}\frac{f(x)}{g(x)}=1$
Then you can say that as $\displaystyle x\to{c}$ $\displaystyle f(x)\sim{g(x)}$