n-0-1-n-2n-1-1-pi-4-2-pi-2-12-3-pi-8-4-pi-2-6-

Question Number 53620 by kaivan.ahmadi last updated on 23/Jan/19

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{2 n + 1} = ?

1 . \frac{π}{4} 2 . \frac{π^{2}}{12} 3 . \frac{π}{8} 4 . \frac{π^{2}}{6}

Commented by turbo msup by abdo last updated on 24/Jan/19

1) \frac{π}{4} i f y o u w a n t m e t h o d s e e t h e

p l a t f o r m .

Commented by kaivan.ahmadi last updated on 24/Jan/19

thank you sir

Commented by turbo msup by abdo last updated on 24/Jan/19

y o u a r e w e l c o m e

Commented by maxmathsup by imad last updated on 26/Jan/19

l e t S (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} x^{2 n + 1} w i t h ∣ x ∣< 1 d u e t o u n i f o r m c o n v e r g e n c e w e h a v e

\frac{d}{d x} S (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} x^{2 n} = \sum_{n = 0}^{\infty} {(- x^{2})}^{n} = \frac{1}{1 + x^{2}} \Rightarrow

S (x) = \int_{0}^{x} \frac{d t}{1 + t^{2}} + c = a r c t a n (x) + c

c = S (o) = 0 \Rightarrow S (x) = a r c t a n x a n d \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} = S (1) = a r c t a n (1) = \frac{π}{4} .

Answered by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19

S_{n} = \frac{1}{1} - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots \infty

G r e g o r y s e r i e s

{t a n}^{- 1} x = \int \frac{d x}{1 + x^{2}}

{t a n}^{- 1} x = \int (1 - x^{2} + x^{4} - x^{6} + \dots) d x

= x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \dots

p u t x = 1 b o t h s i d e

{t a n}^{- 1} (1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \dots

S o S_{n} = {t a n}^{- 1} (1) = \frac{π}{4}

Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19

Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19

Commented by turbo msup by abdo last updated on 24/Jan/19

t h a n k y o u s i r f o r t h i s f o r m u l a s

Commented by tanmay.chaudhury50@gmail.com last updated on 24/Jan/19

m o s t w e l c o m e s i r \dots i h a v e j u s t s h a r e d t h e i n f o r m a t i o n

Commented by kaivan.ahmadi last updated on 24/Jan/19

thank you si \overset{}{r}

Commented by Meritguide1234 last updated on 25/Jan/19

which book is t! <