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- May 16th 2008, 06:15 AMszpengchaor they convergent or not?
- May 16th 2008, 07:12 AMflyingsquirrel
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 AMszpengchaosorry
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 AMDr Zoidburg
- May 16th 2008, 07:51 AMflyingsquirrel
- May 16th 2008, 08:08 AMszpengchaook
so how can u show that sequence tends to 1/n ????

- May 16th 2008, 08:16 AMflyingsquirrel
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 PMMathstud28