For the convergance test you either have to find the integral - if its divergent then the series is divergent, etc, etc.

For the radius of convergence: do the ratio test, L = lim k-->inf |f(k+1)/f(k)|, if L > 1 div, if L < 1, con. absolutely. If L = 1 or lim does not exist then the test is inconclusive.

#8 - I think you can use the ratio test and get |(2n+1)!(2n+1)!/(4n+1)! * (4n)!/(2n)!(2n)!| which cancels to (2n+1)(2n+1) / (4n+1) so then take the lim as n --> inf and if L < 1 then it's convergent.

#10 is the alternating series test - if 0 < an+1 < an and lim n--> inf = 0 then the series is absolutely convergent for all n.

#11 try it again w/ the alternating series test.

#12 the radius of convergence is the ratio test again, then divide each term by n to get -1 < something * x <1. so solving the inequality for x will give you the intervals and plug them into the series to see if it is div or conv at those particular points and that tells you if it's a closed or open interval, ie [x,y], [x,y), (x,y), etc.

#17 - i forget haha

#19 & #21 - seperable diff eq, i forget those too, hahahaa.

good luck. calc 3 sucks hard compared to calc 2.