find a general formula for the sequence. Determine if it converges or diverges a

Question

find a general formula for the sequence. Determine if it converges or diverges and find the limit at which it converges 1){an}= {1,-1/4,1/9,-1/16,1/25...}
2){bn}= {0,3,8,15,24....}
3){cn}= {1,3,1,3...}
4){dn}= {0,1,1,2,2,3,3,4,4,...}

Explanation / Answer

(2n + 1)! = (2n + 1) * (2n) * (2n - 1) * (2n - 2) * ... * 3 * 2 * 1 = (2n + 1) * (2n) * [(2n - 1) * (2n - 2) * ... * 3 * 2 * 1] = (2n + 1) * (2n) * (2n - 1)!. So we have: lim (n-->infinity) (a_n) => lim (n-->infinity) (2n - 1)!/(2n + 1)! = lim (n-->infinity) (2n - 1)!/[(2n + 1) * (2n) * (2n - 1)!] = lim (n-->infinity) 1/[2n(2n + 1)], by canceling (2n - 1)! = lim (n-->infinity) (1/n^2)/[2(2 + 1/n)], by dividing numerator and denominator by n^2 = 0/[2(2 + 0)] = 0. Therefore, the sequence converges to zero. You didn't state what n converges to, but I'll assume 8. 1) Divide numerator and denominator by n^3, getting: (1 / (9 + 1/n^3)) As n->8, n^3->8, and (1/n^3)->0. So as n->8, (1 / (9 + (1/n^3))-> 1/9. Answer to #1: This sequence converges to 1/9. ------------- 2) sqrt((n+9)/(36n+9)) Divide the numerator and denominator inside of the square root sign by n. sqrt( [(1 + (9/n)) / (36 + (9/n))] ) As n->8, (9/n) -> 0, so the limit of this sequence is sqrt(1/36) = 1/6. Answer to #2: this sequence converges to 1/6. --------------- 3) ln(2n^2+3)-ln(n^2+2) This is a basic theorem about logarithms: Theorem) log(x / y) = log(x) - log(y) for any base and for both x and y positive. The logarithm of z is not defined for z 8, (3/n^2) and (2/n^3)->0 Answer to #3: So this converges to ln(2/1) = ln(2) ------------------------ I'm sure that you see the pattern of the technique for this type of problem. As long as division by some power of n gets rid of some parts of these fractions, that's what you should do, but I'll bet that you have already figured that out :D .

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