let-u-n-k-1-n-1-k-k-1-prove-that-u-n-is-convergente-2-find-a-equivalent-of-u-n-when-n-

Question Number 42190 by maxmathsup by imad last updated on 19/Aug/18

l e t u_{n} = \sum_{k = 1}^{n} \frac{{(- 1)}^{k}}{\sqrt{k}}

1) p r o v e t h a t (u_{n}) i s c o n v e r g e n t e

2) f i n d a e q u i v a l e n t o f u_{n} w h e n n \to + \infty

Commented by maxmathsup by imad last updated on 21/Aug/18

1) w e h a v e {l i m}_{n \to + \infty} u_{n} = \sum_{k = 1}^{\infty} \frac{{(- 1)}^{k}}{\sqrt{k}} a n d t h i s s e r i e i s c o n v e r g e n t e

(a l t e r n a t e s e r i e)

2) w e h a v e u_{n} = \sum_{p = 1}^{[\frac{n}{2}]} \frac{1}{\sqrt{2 p}} - \sum_{p = 0}^{[\frac{n - 1}{2}]} \frac{1}{\sqrt{2 p + 1}} = Σ v_{p} - Σ w_{p}

(v_{p}) i s d e c r e a s i n g \Rightarrow \int_{p}^{p + 1} \frac{d x}{\sqrt{2 x}} ⩽ v_{p} ⩽ \int_{p - 1}^{p} \frac{d x}{\sqrt{2 x}} \Rightarrow

\sum_{p = 1}^{[\frac{n}{2}]} \int_{p}^{p + 1} \frac{d x}{\sqrt{2 x}} ⩽ \sum_{p = 1}^{[\frac{n}{2}]} v_{p} ⩽ \sum_{p = 1}^{[\frac{n}{2}]} \int_{p - 1}^{p} \frac{d x}{\sqrt{2 x}} \Rightarrow

\int_{1}^{[\frac{n}{2}] + 1} \frac{d x}{\sqrt{2 x}} ⩽ \sum_{p = 1}^{[\frac{n}{2}]} v_{p} ⩽ \int_{0}^{[\frac{n}{2}]} \frac{d x}{\sqrt{2 x}} \Rightarrow

\frac{1}{\sqrt{2}} {[2 \sqrt{x}]}_{1}^{[\frac{n}{2}] + 1} ⩽ \sum_{p = 1}^{[\frac{n}{2}]} v_{p} ⩽ \frac{1}{\sqrt{2}} {[2 \sqrt{x}]}_{0}^{[\frac{n}{2}]} \Rightarrow \sqrt{2} {\sqrt{[\frac{n}{2}] + 1} - 1} ⩽ Σ v_{p} ⩽ \sqrt{2} \sqrt{[\frac{n}{2}]}

a l s o (w_{p}) i s d e c r e a s i n g \Rightarrow \int_{p}^{p + 1} \frac{d x}{\sqrt{2 x + 1}} ⩽ w_{p} ⩽ \int_{p - 1}^{p} \frac{d x}{\sqrt{2 x + 1}} \Rightarrow

\sum_{p = 1}^{[\frac{n - 1}{2}]} \int_{p}^{p + 1} \frac{d x}{\sqrt{2 x + 1}} ⩽ \sum_{p = 1}^{[\frac{n - 1}{2}]} w_{p} ⩽ \sum_{p = 1}^{[\frac{n - 1}{2}]} \int_{p - 1}^{p} \frac{d x}{\sqrt{2 x + 1}} \Rightarrow

\int_{1}^{[\frac{n - 1}{2}] + 1} \frac{d x}{\sqrt{2 x + 1}} ⩽ \sum_{p = 1}^{[\frac{n - 1}{2}]} w_{p} ⩽ \int_{0}^{[\frac{n - 1}{2}]} \frac{d x}{\sqrt{2 x + 1}} \Rightarrow

{[\sqrt{2 x + 1}]}_{1}^{\frac{[n - 1}{2}] + 1} ⩽ \sum_{p = 1}^{[\frac{n - 1}{2}]} w_{p} ⩽ {[\sqrt{2 x + 1}]}_{0}^{[\frac{n - 1}{2}]} \Rightarrow

\sqrt{2 [\frac{n - 1}{2}] + 2} - \sqrt{3} ⩽ \sum_{p = 1}^{[\frac{n - 1}{2}]} w_{p} ⩽ \sqrt{2 [\frac{n - 1}{2}] + 1} - 1 \Rightarrow

\sqrt{2 [\frac{n - 1}{2}] + 2} + 1 - \sqrt{3} ⩽ \sum_{p = 0}^{[\frac{n - 1}{2}]} w_{p} ⩽ \sqrt{2 [\frac{n - 1}{2}] + 1} - 2 \Rightarrow

\sqrt{2} {\sqrt{[\frac{n}{2}] + 1} - 1} + 2 - \sqrt{2 [\frac{n - 1}{2}] + 1} ⩽ u_{n} ⩽ \sqrt{2} \sqrt{[\frac{n}{2}]} + \sqrt{3} - 1 - \sqrt{2 [\frac{n - 1}{2}] + 1}

\dots b e c o n t i n u e d \dots