1) (a) lim sin(1/n)1/n=exp(lim sin(1/x)/x)=exp [LíHopitalís Rule lim (-1)* (1/x)*(1/x)/sin(1/x)]=exp(0)=1.
(b) Since series 3n/n! converges then lim =0.
( c) lim ln(n2+1)/n1/3=LíHopitalís Rule lim6*x5/6/x6/3(1+1/x2)=0.
( d) lim en*n-1/en-1= L.R. lim 2x*ex*x-x=infinity.
2) (a) integral x-1/3is improper at 0 . Do the calculation to get converges.
(b) x-4/3 is improper at 0, and + and Ė infinity. Calculations yield integral d.n.e.
(c ) x-4/3 is improper at infinity. Calculations give convergence.
(d) e-x*x is improper at infinity and bounded by e-x for x.1, which converges.
3) (a) †series (-1)n2n*n/nn †does not converge absolutely by the root test which also show the term does not go to 0. Series diverges.
(b) series (-1)n/nln(n) converges conditionally by Alt.Series Test but not Abs. Bt the integral test.
(c ) series (-1)n(n)1/2/n2+1 converges absolutely bu the limit comparison to 1/n3/2
(d) series ln(n)/n3/2 converges by the limit comparison to 1/n5/4
4) Start with the geometric series . Replace x by -x2 . integrate the series term by term to get a power for the arctan(x). †Replace x by x2.
Convergence is |x|<=1(Alternating Series at |x|=1) †We end up with the series arctan(x2)=series(-1)n*x4n+2/2n+1.
5) ratio test gives series xn/n3 converges in |x|<1 . Integral test give abs. Convergence at x=1 and x=-1.