Question

Can anyone show me how this converges, please? (2n-1)!/(2n+1)!

Answer 1

?
I suppose,
\[\sum^b_{n=0}\frac{(2n+1)!}{(2n+1)!}=\sum^b_{n=0}1=b\]

Answer 2

Oh I see lol I made a sign mistake I give up I got to go to bed lol

Answer 3

(2n-1)!/(2n+1)(2n)(2n-1)!

Answer 4

so your function actually is
1/(2n+1)(2n) = 1/ (4n^2 +2n)
use any test to show that this converges.

Answer 5

k bro