Question

here's one that seems a little bit tough: Test it's convergence (absolutely, conditionally, or divergent)
\[\sum_{n=1}^{\infty}{(-1)^n\frac{2^n*n!}{5*8*11*...*(3n+2)}}\]

Answer 1

its*

Answer 2

use ratio test

Answer 3

yeah that's what I thought.... but I hate the ratio test :-(

Answer 4

integral test and root test are out of the question. Haven't tried the simple limit test; don't know if it applies here.

Answer 5

use the ratio test...it really isn't that bad

Answer 6

simple limit test gives me a negative infinity answer... so it is divergent :-D

Answer 7

or am I mistaken? what does the ratio test tell me?

Answer 8

looks like it converges to me

Answer 9

http://www.wolframalpha.com/input/?i=sum+from+1+to+infinity+%28%28-1%29%5En+*+%28%282%5En*n%21%29%2F%285*8*11*...*%283n%2B2%29%29%29%29

Answer 10

the magic of computing!

Answer 11

I don't think that is correct

Answer 12

Mathematica version 8 says it converges

Answer 13

I am willing to believe Mathematica 8, but does it have like a working out or a proof?

Answer 14

formatting for the copy is a little messed up ... but

In[2]:= SumConvergence[((-1)^n*2^n*Factorial[n])/(\!\(
\*UnderoverscriptBox[\(\[Product]\), \(k = 1\), \(n\)]\(3 k\)\)+2),n]
Out[2]= True

Answer 15

the ratio test works...gives me 2/3 which is less than 1

Answer 16

I better try it.... I'm just so scared of the 2^n and the 6*8*11*.....

Answer 17

\[\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(-1)^{n+1}\frac{2^{n+1}*(n+1)!}{5*8*11*...*(3n+2)(3n+5)}}{(-1)^n\frac{2^n*n!}{5*8*11*...*(3n+2)}}\right|\]

\[=\frac{2^{n+1}*(n+1)!}{5*8*11*...*(3n+2)(3n+5)}{\frac{5*8*11*...*(3n+2)}{2^n*n!}}\]

\[=\frac{2(n+1)}{3n+5}\to\frac{2}{3}<1\]

Answer 18

doesn't look so bad... but why would wolfram be wrong :-(... I trusted him so much.

Answer 19

not sure..since it is the same company that makes mathematica

Answer 20

it's not the first time I've seen wolfram wrong

Answer 21

what else does it do wrong?

Answer 22

I don't remember the specific problem