r they convergent or not?

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May 16th 2008, 06:15 AM
szpengchao

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r they convergent or not?

http://www.mathhelpforum.com/math-he...1&d=1210947307
May 16th 2008, 07:12 AM
flyingsquirrel

Hi

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

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

The only other way I can think of, off-hand, is that $g$ approximates to $f$ .
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, $\sum \frac{n^3+(-1)^n8n^2+1}{n^4+(-1)^{n+1}n^2}$ 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

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

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

Quote:

Originally Posted by szpengchao

so how can u show that sequence tends to 1/n ????

I actually just put a post up about this..

If $\lim_{x\to{c}}\frac{f(x)}{g(x)}=1$

Then you can say that as $x\to{c}$ $f(x)\sim{g(x)}$

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