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.33. To determine if this sequence converges, we can find the limit as approaches infinity. Therefore, the sequence converges and its limit is .35. To determine if this sequence converges, we can find the limit as approaches infinity. Therefore, the sequence converges and its limit is .37. To determine if this sequence converges, we can find the limit as approaches infinity.$\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 \