# determining absolute or conditional convergence

• Mar 25th 2010, 05:17 PM
oblixps
determining absolute or conditional convergence
This is a question that was on my quiz.

sum (n=1 to infinite) of (-1)^(n+1) (n!)^2 / (2n)!

i did the alternating series test and showed that absolute value of each term in the series approaches 0. then i showed that the absolute value of each term decreases as n increases. thus this alternating series converges. Then i compared the absolute value of this series to (1/2)^n and found that since this series is less than (1/2)^n and (1/2)^n converges, then this smaller series must also converge due to the direct comparison test. so my answer to this question is that this series converges absolutely.

but my teacher marked this question wrong and put a red line through my answer and my work. did i do something wrong?
• Mar 25th 2010, 06:49 PM
skeeter
Quote:

Originally Posted by oblixps
This is a question that was on my quiz.

sum (n=1 to infinite) of (-1)^(n+1) (n!)^2 / (2n)!

i did the alternating series test and showed that absolute value of each term in the series approaches 0. then i showed that the absolute value of each term decreases as n increases. thus this alternating series converges. Then i compared the absolute value of this series to (1/2)^n and found that since this series is less than (1/2)^n and (1/2)^n converges, then this smaller series must also converge due to the direct comparison test. so my answer to this question is that this series converges absolutely.

but my teacher marked this question wrong and put a red line through my answer and my work. did i do something wrong?

familiar with the ratio test ?

$\lim_{n \to \infty} \left| \frac{[(n+1)!]^2}{[2(n+1)]!} \cdot \frac{(2n)!}{(n!)^2}\right| =$

$\lim_{n \to \infty} \left| \frac{[(n+1)n!]^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2}\right| =$

$\lim_{n \to \infty} \left| \frac{(n+1)^2(n!)^2}{(2n+2)!} \cdot \frac{(2n)!}{(n!)^2}\right| =$

$\lim_{n \to \infty} \left| \frac{(n+1)^2}{(2n+2)(2n+1)}\right| = \frac{1}{4} < 1$

the series converges absolutely
• Mar 26th 2010, 10:12 PM
oblixps
ah so i was correct then. thank you for verifying! (Nod)
• Mar 27th 2010, 04:17 AM
skeeter
Quote:

Originally Posted by oblixps
ah so i was correct then. thank you for verifying! (Nod)

your conclusion was correct, but how did you show that $\frac{(n!)^2}{(2n)!} < \frac{1}{2^n}$ ... ? maybe that's why you did not get credit.