Follow Math Help Forum on Facebook and Google+
Hi For the first one, you can try to show that for . For the second one, the ratio test may be helpful. Not completely off-topic question : If do English speaking people say that is an approximation of ? Is there another way of saying this ?
Last edited by flyingsquirrel; May 16th 2008 at 09:10 AM.
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?
Originally Posted by flyingsquirrel Not completely off-topic question : If do English speaking people say that is an approximation of ? Is there another way of saying this ? The only other way I can think of, off-hand, is that approximates to .
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, can't be convergent) Anyway, the comparison test can also be used to show the divergence.
so how can u show that sequence tends to 1/n ????
Originally Posted by szpengchao so how can u show that sequence tends to 1/n ???? You can get it by showing that tends to 1 as approaches . (divide the numerator and the denominator by )
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 Then you can say that as
View Tag Cloud