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**1) ****(a) lim sin(1/n) ^{1/n}=exp(lim
sin(1/x)/x)=exp [L’Hopital’s Rule lim (-1)* (1/x)*(1/x)/sin(1/x)]=exp(0)=1.**

**(b) Since series 3 ^{n}/n! converges then lim =0.**

**( c) lim ln(n ^{2}+1)/n^{1/3}=L’Hopital’s Rule lim6*x^{5/6}/x^{6/3}(1+1/x^{2})=0.**

**( d) lim e ^{n*n}-1/e^{n}-1= L.R. lim 2x*e^{x*x-x}=infinity.**

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**2) ****(a) integral x ^{-1/3}is improper at
0 . Do the calculation to get converges.**

**(b) x ^{-4/3} is improper at 0, and + and – 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) ^{n}2^{n*n}/n^{n} 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}/n^{2}+1 converges
absolutely bu the limit comparison to 1/n^{3/2}**

**(d) series ln(n)/n ^{3/2} converges by the limit comparison to 1/n^{5/4}**

**4) ****Start with the geometric series . Replace x
by -x ^{2} . integrate the series term by term to get a power for the arctan(x).
Replace x by x^{2}.**

**Convergence is |x|<=1(Alternating Series at |x|=1) We end up with the series arctan(x ^{2})=series(-1)^{n}*x^{4n+2}/2n+1.**

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**5) ratio test gives series x ^{n}/n^{3} converges in
|x|<1 . Integral test give abs. Convergence at x=1 and x=-1. **

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