Given the following sequences a_n = sigma_k = 1^n 1/k, b_n sigma_k = 1^n k^2, c_

Question

Given the following sequences a_n = sigma_k = 1^n 1/k, b_n sigma_k = 1^n k^2, c_n = sigma_k = 1^n, 1/k^2, d_n = sigma_k = 1^n k, e_n = sigma_k = 1^n 1/e^k, f_n = [n/2], g_n = [n/2]. Compute lim_n rightarrow infinity a_n/c_n. Find lim_n rightarrow infinity b_n/n^2. Find lim _n rightarrow infinity d_n/n^3. Find lim_n rightarrow infinity c_n/log n. Find lim_n rightarrow infinity f_n/n log n. Find lim_n rightarrow infinity g_n/f_n. As n rightarrow infinity, which sequence (s) is (are) of O(1)? As n rightarrow infinity, which sequence (s) is (are) of (log n)? As n rightarrow infinity, which sequence (s) is (are) of O(n log n)? As n rightarrow infinity, which sequence (s) is (are)of O(n)? As n rightarrow infinity, which sequence (s) is (are) of O (n^2) As n rightarrow infinity, which sequence (s) is (are) of O (n^3)? As n rightarrow infinity, which sequence (s) is (are) of O (n^4)?

Explanation / Answer

Part (a)

en = (1/e) + (1/e2) + (1/e3) + ...... + (1/en) is sum to n terms of a GP with a = 1/e and r = 1/e. Sum to n terms of a GP with first term a and common ratio r is given Sn = a(1 - rn)/(1 - r). So, an = (1/e){1- (1/e)n)/{1 - (1/e)} = [(1/e){(en - 1)/en}]{e/(e - 1)} = (en - 1)/{en(en - 1)} = {1 - (1/en)}/(e - 1) [Dividing both numerator and denominator by en].

As n tends to infinity, (1/en) tends to 0 and hence the required limit = 1/(e - 1) ANSWER

Part c

bn = 12 + 22 + ..... + n2 = sum of squares of first n natural numbers = n(n + 1)(2n + 1)/6 = n3{1 +(1/n)}{2 + (1/n)}/6.

Hence, bn/n2 = n{1 +(1/n)}{2 + (1/n)}/6. This clearly tends to infinity as n tends to infinity.

=> limit does not exist ANSWER  

[If however, it were  bn/n3= {1 +(1/n)}{2 + (1/n)}/6, the limit as n tends to infinity wouls have been,  {1 +(0)}{2 + (0)}/6 = 2/6 = 1/3 since as n tends to infinity (1/n) tends to zero.]

Part (d)

dn = 1 + 2 + ..... + n = sum of first n natural numbers = n(n + 1)/2 = n2 {1 +(1/n)}/2 and hence dn/n2 = {1 +(1/n)}/2 which tends to (1 + 0)/2 = 1/2 as n tends to infinity [since as n tends to infinity (1/n) tends to zero.]

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