Thread: r they convergent or not?

1. r they convergent or not?

2. 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 ?

3. 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?

4. 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 $\displaystyle f$.

5. 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.

6. ok

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

7. 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 $\displaystyle n^4$)

8. 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 $\displaystyle x\to{c}$ $\displaystyle f(x)\sim{g(x)}$