1. ## limits

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

2. Originally Posted by jeph
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
It is zero.

3. ooh, ooh! I know this one!

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!

4. Originally Posted by Nerd
ooh, ooh! I know this one!

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!
Or you can just say that,

|2/n^20(n+1)^20| <= 1/n

And lim n-->+oo (1/n)=0

Then certainly that one goes to 0 as well.

5. 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?

6. Originally Posted by jeph
is that the direct comp test?
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.

7. oh...i dont think we ever learned the squeeze thereom ._.