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Question Number 37333 by math khazana by abdo last updated on 12/Jun/18

l e t f (x) = \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n^{3}}

1) s t u d y t h e c o n v e r g e n c e o f t h i s s e r i e

2) p r o v e t h a t \int_{0}^{π} f (x) d x = 2 \sum_{n = 1}^{\infty} \frac{1}{{(2 n - 1)}^{4}}

3) p r o v e t h a t \forall x \in \in R f^{^{'}} (x) = \sum_{n = 1}^{\infty} \frac{c o s (n x)}{n^{2}}

4) p r o v e t h a t \int_{0}^{\frac{π}{2}} (\sum_{n ⩾ 1} \frac{c o s (n x)}{n^{2}}) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{2}}

Commented by prof Abdo imad last updated on 17/Jun/18

1) t h e u n i f o r m c o n v e r g e n c e o f t h i s s e r i e i s

a s s u r e d b e c a u s e \forall x \in R ∣ \frac{s i n (n x)}{n^{3}} ∣⩽ \frac{1}{n^{3}} a n d

Σ \frac{1}{n^{3}} i s c o m v e r g e n t

2) \int_{0}^{π} f (x) d x = \int_{0}^{π} \sum_{n = 1}^{\infty} \frac{s i n (n x)}{n^{3}} d x

= \sum_{n = 1}^{\infty} \frac{1}{n^{3}} \int_{0}^{π} s i n (n x) d x b u t

\int_{0}^{π} s i n (n x) d x = {[- \frac{1}{n} c o s (n x)]}_{0}^{π} = \frac{1}{n} (1 - {(- 1)}^{n})

\Rightarrow \int_{0}^{π} f (x) d \overset{}{x} = \sum_{n = 1}^{\infty} \frac{1}{n^{4}} (1 - {(- 1)}^{n})

= 2 \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{4}} =_{n + 1 = p} 2 \sum_{p = 1}^{\infty} \frac{1}{{(2 p - 1)}^{4}}

Commented by prof Abdo imad last updated on 17/Jun/18

3) l e t f_{n} (x) = \frac{s i n (n x)}{n^{3}} w e h a v e Σ f_{n} c . u n i f .

a n d Σ f_{n}^{^{'}} c o n v . u n i f . \Rightarrow f^{^{'}} (x) = Σ f_{n}^{^{'}} (x)

= \sum_{n = 1}^{\infty} \frac{n c o s (n x)}{n^{3}} = \sum_{n = 1}^{\infty} \frac{c o s (n x)}{n^{2}}

4) d u e t o u n i f o r m c o n v . w e h a v e a l s o

\int_{0}^{\frac{π}{2}} (\sum_{n = 1}^{\infty} \frac{c o s (n x)}{n^{2}}) = \sum_{n = 1}^{\infty} \frac{1}{n^{2}} \int_{0}^{\frac{π}{2}} c o s (n x) d x

= \sum_{n = 1}^{\infty} \frac{1}{n^{2}} {[\frac{1}{n} s i n (n x)]}_{0}^{\frac{π}{2}}

= \sum_{n = 1}^{\infty} \frac{1}{n^{3}} s i n (n \frac{π}{2}) = \sum_{n = 0}^{\infty} \frac{1}{{(2 n + 1)}^{3}} s i n ((2 n + 1) \frac{π}{2})

= \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{{(2 n + 1)}^{3}} .