let-S-n-p-k-1-n-1-k-p-find-a-equivalent-of-S-n-p-when-n-p-integr-1-

Question Number 42098 by abdo.msup.com last updated on 17/Aug/18

l e t S_{n, p} = \sum_{k = 1}^{n} \frac{1}{\sqrt{k + p}}

f i n d a e q u i v a l e n t o f S_{n, p} w h e n n \to + \infty

p i n t e g r ⩾ 1 .

Commented by maxmathsup by imad last updated on 18/Aug/18

t h e s e q u e n c e {(\frac{1}{\sqrt{k}})}_{k ⩾ 1} i s d e c r e a s i n g \Rightarrow \int_{k}^{k + 1} \frac{d t}{\sqrt{t + p}} ⩽ \frac{1}{\sqrt{k + p}} ⩽ \int_{k - 1}^{k} \frac{d t}{\sqrt{t + p}} \Rightarrow

\sum_{k = 1}^{n} \int_{k}^{k + 1} \frac{d t}{\sqrt{t + p}} ⩽ \sum_{k = 1}^{n} \frac{1}{\sqrt{k + p}} ⩽ \sum_{k = 1}^{n} \int_{k - 1}^{k} \frac{d t}{\sqrt{t + p}} \Rightarrow

\int_{1}^{n + 1} \frac{d t}{\sqrt{t + p}} ⩽ S_{n, p} ⩽ \int_{0}^{n} \frac{d t}{\sqrt{t + p}} \Rightarrow {[2 \sqrt{t + p}]}_{1}^{n + 1} ⩽ S_{n, p} ⩽ {[2 \sqrt{t + p}]}_{0}^{n} \Rightarrow

2 \sqrt{n + 1 + p} - 2 \sqrt{p + 1} ⩽ S_{n, p} ⩽ 2 \sqrt{n + p} - 2 \sqrt{p} \Rightarrow

\frac{2 \sqrt{n + 1 + p} - 2 \sqrt{p + 1}}{2 \sqrt{n + p}} ⩽ \frac{S_{n, p}}{2 \sqrt{n + p}} ⩽ \frac{2 \sqrt{n + p} - 2 \sqrt{p}}{2 \sqrt{n + p}} \Rightarrow

\sqrt{1 + \frac{1}{n + p}} - \sqrt{\frac{p + 1}{n + p}} ⩽ \frac{S_{n, p}}{2 \sqrt{n + p}} ⩽ 1 - \sqrt{\frac{p}{n + p}} b u t

{l i m}_{n \to + \infty} \sqrt{1 + \frac{1}{n + p}} - \sqrt{\frac{p + 1}{n + 1}} = 1 a n d {l i m}_{n \to + \infty} 1 - \sqrt{\frac{p}{n + p}} = 1 \Rightarrow

S_{n, p} \sim 2 \sqrt{n + p} (n \to + \infty a n d p f i x e d)