how do you solve something like this where there is a high exponent polynomial thing like

lim_n->oo 2 / n^20(n+1)^20

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- May 8th 2007, 01:13 PMjephlimits
how do you solve something like this where there is a high exponent polynomial thing like

lim_n->oo 2 / n^20(n+1)^20 - May 8th 2007, 01:19 PMThePerfectHacker
- May 8th 2007, 01:29 PMNerd
ooh, ooh! I know this one! :D

What you got here is a finite number basically divided by infinity^40 which is still just infinity. Any finite number divided by an inifinitly huge one is 0! - May 8th 2007, 01:37 PMThePerfectHacker
- May 8th 2007, 02:47 PMjeph
is that the direct comp test?

the question was sumation n=1 oo 2^n / n^20. i used the ratio test to get 2/n^20(n+1)^20.

on the sheet of paper i have it says 0<b_n<=a_n will diverge

wouldnt 1/n be b_n in this case?

1/n is bigger than 2/n^20(n+1)^20 so it wouldnt work...

edit:

oh wait...it = 0 so it converges....maybe i did something wrong then? because the series is supposed to diverge...

can you show me how to solve sumation n=1 oo 2^n / n^20 to show that it diverges using the ratio test? - May 8th 2007, 03:45 PMThePerfectHacker
Almost! I used the "squeeze theorem" in disguise.

**Theorem:**If {a_n} and {b_n} are convergent sequences to the same limit L and {c_n} is a sequence such that:

a_n <= c_n <= b_n for sufficiently large n

Then {c_n} is convergent to L as well.

**Theorem:**If {a_n} is a convergent sequence to 0 and {c_n} is a sequence such that:

|c_n|<=a_n for sufficiently large n

Then {c_n} is convergent to 0.

**Proof:**We can write |c_n|<=a_n differently:

-a_n<=c_n<=a_n

But the limit of a_n is 0 therefore the limit of -a_n is also 0.

Hence by above theorem the limit c_n exists and equals 0. - May 8th 2007, 07:10 PMjeph
oh...i dont think we ever learned the squeeze thereom ._.