What is the asymptotic behavior of the sequence: s n </msub> = <munderover> &

aniawnua

aniawnua

Answered question

2022-05-22

What is the asymptotic behavior of the sequence:
s n = k = 1 n k 1 / 4

Answer & Explanation

Mihevcekd

Mihevcekd

Beginner2022-05-23Added 7 answers

Let k 1. Since the function x x 1 / 4 is increasing, ( k 1 ) 1 / 4 x 1 / 4 k 1 / 4 for every k 1 x k. Integrating this double inequality yields
( k 1 ) 1 / 4 k 1 k x 1 / 4 d x k 1 / 4 .
Summing these from k = 1 to k = n yields
s n 1 = k = 0 n 1 k 1 / 4 0 n x 1 / 4 d x k = 1 n k 1 / 4 = s n .
Since the integral is 4 5 n 5 / 4 and s n 1 = s n n 1 / 4 , this yields, for every n 1
4 5 n 5 / 4 s n 4 5 n 5 / 4 + n 1 / 4 .
Note finally that n 1 / 4 = o ( n 5 / 4 ) hence a (much weakened) version of this is s n 4 5 n 5 / 4

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