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Question Number 33847 by prof Abdo imad last updated on 26/Apr/18

l e t g i v e a s e q u e n c e o f r e a l n u m b e t s p o s i t i f

{(a_{i})}_{1 ⩽ i ⩽ n}

1) p r o v e t h a t {(\sum_{i = 1}^{n} a_{i})}^{2} ⩽ n \sum_{i = 1}^{n} a_{i}^{2}

2) l e t p u t H_{n} = \sum_{k = 1}^{n} \frac{1}{k} a n d w_{n} = \frac{H_{n}^{2}}{n}

p r o v e t h a t t h e s e q u e n c e w_{n} i s c o n v e r g e n t .

Commented by prof Abdo imad last updated on 27/Apr/18

f o r a l l s e q u e n c e s o f r e a l s n u m b e r s p o s i t i f s

{(a_{i})}_{1 ⩽ i ⩽ n} a n d {(b_{i})}_{1 ⩽ i ⩽ n} w e h a v e

\sum_{i = 1}^{n} a_{i} b_{i} ⩽ {(\sum_{i = 1}^{n} a_{i}^{2})}^{\frac{1}{2}} {(\sum_{i = 1}^{n} b_{i}^{2})}^{\frac{1}{2}} (h o l d e r i n e q u a l i t y)

l e t t a k e b_{i} = 1 \forall i \in [[1, n]] \Rightarrow \sum_{i = 1}^{n} a_{i} ⩽ \sqrt{n} {(\sum_{i = 1}^{n} a_{i}^{2})}^{\frac{1}{2}} \Rightarrow

{(\sum_{i = 1}^{n} a_{i})}^{2} ⩽ n (\sum_{i = 1}^{n} a_{i}^{2})

2) l e t t a k e a_{i} = \frac{1}{i} \forall i \in [1, n] \Rightarrow {(\sum_{i = 1}^{n} \frac{1}{i})}^{2} ⩽ n \sum_{i = 1}^{n} \frac{1}{i^{2}}

\Rightarrow H_{n}^{2} ⩽ n \sum_{i = 1}^{n} \frac{1}{i^{2}} \Rightarrow \frac{H_{n}^{2}}{n} ⩽ \sum_{i = 1}^{n} \frac{1}{i^{2}} \Rightarrow

w_{n} ⩽ \sum_{i = 1}^{n} \frac{1}{i^{2}} R i e m a n s e r i e c o n v e r g e n t s o

(w_{n}) i s c o n v e r g e n t .