Question

absolute and conditional convergence

Answer 1

one of those series tests *faints*

Answer 2

simplify 2n-1 / 2n-1 ! = 1/2n-2!! * -1 ^ n-1.
you get 1/2*4*6*2n-2 now, do alternating series test:
a n < a n+1
lim n-> infinity a=0
first one is good
second is good
check for absolute convergence though

Answer 3

it's absolute convergent if you use the ratio test

Answer 4

the limit turns out to be 1/2, which is less than 1, so it's absolute convergent, hence, the series also converges

Answer 5

\[

\sum _{n=1}^{\infty } \frac{(-1)^{n-1} (2 n)!}{2^n n! (2
n-1)!}=\frac{1}{\sqrt{e}}\\
\sum _{n=1}^{\infty } \frac{(2 n)!}{2^n n! (2 n-1)!}=\sqrt{e}
\]

Answer 6

Try to show the above.

Answer 7

i got this by ratio test but i dont know how to proceed.\[\lim_{n \rightarrow \infty}\frac{ 1\times3\times5...(2n+1) }{ (2n+1)! }\times \frac{ (2n-1)! }{ 1\times3\times5...(2n-1) }\]