do these converge:
sum from n=1 to infinty of [ (2^n)(2n)! ] / [(9^n)(n!)^2]
and
sum from n=1 to infinity of (n-1) / [(n^2 + 2) (n^2 + 1)^(1/4)]
2007-12-11 02:35:36 · 8 answers · asked by dposters 1 in Science & Mathematics ➔ Mathematics
The first sum converges to 2 and the second sum converges by Ermakoff's test, although I think it converges very slowly to what, I don't know.
2007-12-11 03:09:19 · answer #1 · answered by 1,1,2,3,3,4, 5,5,6,6,6, 8,8,8,10 6 · 0⤊ 0⤋
1)For every n,
a_n = [ (2^n)(2n)! ] / [(9^n)(n!)^2] = (2/9)^n * (1 *2 *...2n)/(1 *2 *....n)^2 = (2/9)^n * (n+1) (n+2)....2n = (n+1)(2/9) * (n +2) *(2/9) * 2n * (2/9) = Product (k=1, n) (n +k) * (2/9). For k >= 4 all the factor of the product are > 1 and are unbpounded. So a_n --> oo, which implies Sum a_n --> oo. Diverges
2) For every n, a_n = (n-1) / [(n^2 + 2) (n^2 + 1)^(1/4)] < (n-1)/(n^2 * (n^2)^(1/4)) = (n -1)(n^2 * n^(1/2)) = (n -1)/(n^(5/2)) < n/(n^5/2) = 1/n^(3/2)
Since 3/2 >1, it follows Sum 1/n^(3/2) converges. By comparisson, we conclude your series converges
2007-12-11 03:33:08 · answer #2 · answered by Steiner 7 · 0⤊ 1⤋
They will get infinitely close but never converge.
2007-12-11 02:40:58 · answer #3 · answered by Brewspy 4 · 0⤊ 1⤋
So confusing!!!, can you just write the questions a bit further away from each other?
2007-12-11 02:42:00 · answer #4 · answered by Nai 4 · 0⤊ 2⤋
the bonus legham does not exist, it is sausage meat wrapped in a lie. penguins know this...that is why they dont wear red! paaaancaaakes! cried the baboon.
2007-12-11 02:39:14 · answer #5 · answered by R-T. 4 · 1⤊ 2⤋
no definitely they wont converge
2007-12-11 02:39:17 · answer #6 · answered by saravanacbe 2 · 0⤊ 1⤋
YOU are too CHEAP
2007-12-11 02:38:58 · answer #7 · answered by Anonymous · 1⤊ 2⤋
NOPE!!!!
2007-12-11 02:38:19 · answer #8 · answered by erin_foss8191@sbcglobal.net 3 · 0⤊ 1⤋