Convergence and Divergence Which of the sequences a_(n)) in Exercises 31-100 converge, and which diverge? Find the limit of each convergent sequence. a_(n)=2+(0.1)^n 33 a_(n)=(1-2n)/(1+2n) 35 a_(n)=(1-5n^4)/(n^4)+8n^(3) 37 a_(n)=(n^2-2n+1)/(n-1) 39 a_(n)=1+(-1)^n 32. a_(n)=(n+(-1)^n)/(n) 34. a_(n)=(2n+1)/(1-3sqrt (n)) 36 a_(n)=(n+3)/(n^2)+5n+6 38. a_(n)=(1-n^3)/(70-4n^2) 40. a_(n)=(-1)^n(1-(1)/(n))

Let's analyze each sequence to determine if it converges or diverges, and find the limit of any convergent sequences.<br /><br />33. $a_{n}=\frac {1-2n}{1+2n}$<br /><br />To determine if this sequence converges, we can find the limit as $n$ approaches infinity. <br /><br />$\lim_{{n \to \infty}} \frac{1-2n}{1+2n} = \lim_{{n \to \infty}} \frac{-2n-1}{2n+1} = \lim_{{n \to \infty}} \frac{-2 - \frac{1}{n}}{2 + \frac{1}{n}} = \frac{-2}{2} = -1$<br /><br />Therefore, the sequence converges and its limit is $-1$.<br /><br />35. $a_{n}=\frac {1-5n^{4}}{n^{4}+8n^{3}}$<br /><br />To determine if this sequence converges, we can find the limit as $n$ approaches infinity.<br /><br />$\lim_{{n \to \infty}} \frac{1-5n^{4}}{n^{4}+8n^{3}} = \lim_{{n \to \infty}} \frac{-5n^{4}-1}{n^{4}+8n^{3}} = \lim_{{n \to \infty}} \frac{-5 - \frac{1}{n^{4}}}{1 + \frac{8}{n}} = \frac{-5}{1} = -5$<br /><br />Therefore, the sequence converges and its limit is $-5$.<br /><br />37. $a_{n}=\frac {n^{2}-2n+1}{n-1}$<br /><br />To determine if this sequence converges, we can find the limit as $n$ approaches infinity.<br /><br />$\lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \to \infty}} \frac{n^{2}-2n+1}{n-1} = \lim_{{n \