calculate-n-0-1-n-4n-1-1-1-5-1-9-1-13-

Question Number 53795 by maxmathsup by imad last updated on 25/Jan/19

c a l c u l a t e \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = 1 - \frac{1}{5} + \frac{1}{9} - \frac{1}{13} + \dots .

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

l e t S (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} x^{4 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^{4 n} = \sum_{n = 0}^{\infty} {(- x^{4})}^{n} = \frac{1}{1 + x^{4}} \Rightarrow S (x) = \int_{0}^{x} \frac{d t}{t^{4} + 1} + c

c = S (0) = 0 \Rightarrow S (x) = \int_{0}^{x} \frac{d t}{t^{4} + 1} a n d \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = S (1) = \int_{0}^{1} \frac{d t}{t^{4} + 1}

l e t d e c o m p o s e F (t) = \frac{1}{t^{4} + 1} \Rightarrow F (t) = \frac{1}{{(t^{2} + 1)}^{2} - 2 t^{2}} = \frac{1}{(t^{2} + \sqrt{2} t + 1) (t^{2} - \sqrt{2} t + 1)}

= \frac{a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{c t + d}{t^{2} - \sqrt{2} t + 1}

F (- t) = F (t) \Rightarrow \frac{- a t + b}{t^{2} - \sqrt{2} t + 1} + \frac{- b t + c}{t^{2} + \sqrt{2} t + 1} = F (t) \Rightarrow c = - a a n d d = b \Rightarrow

F (t) = \frac{a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{- a t + b}{t^{2} - \sqrt{2} t + 1}

F (0) = 1 = 2 b \Rightarrow b = \frac{1}{2}

F (1) = \frac{1}{2} = \frac{a + b}{2 + \sqrt{2}} + \frac{- a + b}{2 - \sqrt{2}} \Rightarrow \frac{1}{2} = \frac{(2 - \sqrt{2}) (a + \frac{1}{2}) + (2 + \sqrt{2}) (- a + \frac{1}{2})}{2} \Rightarrow

- 2 \sqrt{2} a + \frac{1}{2} (4) = 1 \Rightarrow - 2 \sqrt{2} a = - 1 \Rightarrow a = \frac{1}{2 \sqrt{2}} \Rightarrow

F (t) = \frac{\frac{t}{2 \sqrt{2}} + \frac{1}{2}}{t^{2} + \sqrt{2} t + 1} + \frac{- \frac{t}{2 \sqrt{2}} + \frac{1}{2}}{t^{2} - \sqrt{2} t + 1} = \frac{1}{2 \sqrt{2}} {\frac{t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} + \frac{- t + \sqrt{2}}{t^{2} - \sqrt{2} t + 1}} \Rightarrow

\int_{0}^{1} \frac{d t}{t^{4} + 1} = \frac{\sqrt{2}}{4} {\int_{0}^{1} \frac{t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t - \int_{0}^{1} \frac{t - \sqrt{2}}{t^{2} - \sqrt{2} t + 1}}

= \frac{\sqrt{2}}{4} {\frac{1}{2} \int_{0}^{1} \frac{2 t + \sqrt{2} + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} d t - \frac{1}{2} \int_{0}^{1} \frac{2 t - \sqrt{2} - \sqrt{2}}{t^{2} - \sqrt{2} t + 1} d t}

= \frac{\sqrt{2}}{8} {{[l n (t^{2} + \sqrt{2} t + 1)]}_{0}^{1} - {[l n (t^{2} - \sqrt{2} t + 1)]}_{0}^{1} + \sqrt{2} \int_{0}^{1} \frac{d t}{t^{2} + \sqrt{2} t + 1}

+ \sqrt{2} \int_{0}^{1} \frac{d t}{t^{2} - \sqrt{2} t + 1}}

= \frac{\sqrt{2}}{8} {l n (2 + \sqrt{2}) - l n (2 - \sqrt{2})} + \frac{1}{4} {\int_{0}^{1} \frac{d t}{t^{2} + \sqrt{2} t + 1} + \int_{0}^{1} \frac{d t}{t^{2} - \sqrt{2} t + 1}} b u t

\int_{0}^{1} \frac{d t}{t^{2} + \sqrt{2} t + 1} = \int_{0}^{1} \frac{d t}{t^{2} + 2 \frac{\sqrt{2}}{2} t + \frac{1}{2} + \frac{1}{2}} = \int_{0}^{1} \frac{d t}{{(t + \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}}

=_{t + \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}}} \int_{1}^{1 + \sqrt{2}} \frac{d u}{\sqrt{2} \frac{1}{2} (1 + u^{2})} = \sqrt{2} \int_{1}^{1 + \sqrt{2}} \frac{d u}{u} = \sqrt{2} {a r c t a n (\sqrt{2} + 1) - \frac{π}{4}}

\int_{0}^{1} \frac{d t}{t^{2} - \sqrt{2} t + 1} = \int_{0}^{1} \frac{d t}{{(t - \frac{1}{\sqrt{2}})}^{2} + \frac{1}{2}} =_{t - \frac{1}{\sqrt{2}} = \frac{u}{\sqrt{2}}} \int_{- 1}^{\sqrt{2} - 1} \frac{d u}{\sqrt{2} \frac{1}{2} (1 + u^{2})}

= \sqrt{2} {a r c t a n (\sqrt{2} - 1) + \frac{π}{4}} \Rightarrow

\int_{0}^{1} \frac{d t}{t^{4} + 1} = \frac{\sqrt{2}}{8} l n (\frac{2 + \sqrt{2}}{2 - \sqrt{2}}) + \frac{\sqrt{2}}{4} {a r c t a n (\sqrt{2} + 1) + a r c t a n (\sqrt{2} - 1)} b u t

a r c t a n (\sqrt{2} - 1) = a r c t a n (\frac{1}{\sqrt{2} + 1}) = \frac{π}{2} - a r c t a n (\sqrt{2} + 1) \Rightarrow

\int_{0}^{1} \frac{d t}{t^{4} + 1} = \frac{\sqrt{2}}{8} l n (\frac{2 + \sqrt{2}}{2 - \sqrt{2}}) + \frac{π \sqrt{2}}{8} \Rightarrow \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n + 1} = \frac{\sqrt{2}}{8} l n (\frac{2 + \sqrt{2}}{2 - \sqrt{2}}) + \frac{π \sqrt{2}}{8} .

Answered by Smail last updated on 26/Jan/19

\frac{1}{1 - x} = \underset{n = 0}{\sum^{\infty}} x^{n}

S o \frac{1}{1 + x^{4}} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} x^{4 n}

l e t p (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} = \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{x} t^{4 n} d t = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{4 n + 1} x^{4 n + 1}

p (1) = \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{4 n + 1}

p (x) = \int_{0}^{x} \frac{d t}{1 + t^{4}} = \int_{0}^{x} \frac{d t}{(t^{2} + \sqrt{2} t + 1) (t^{2} - \sqrt{2} t + 1)}

\frac{1}{1 + t^{4}} = \frac{a t + b}{t^{2} + \sqrt{2} t + 1} + \frac{c t + d}{t^{2} - \sqrt{2} t + 1}

a = - c = \frac{\sqrt{2}}{4}; d = b = \frac{1}{2}

p (x) = \frac{\sqrt{2}}{4} \int_{0}^{x} (\frac{t + \sqrt{2}}{t^{2} + \sqrt{2} t + 1} - \frac{t - \sqrt{2}}{t^{2} - \sqrt{2} t + 1}) d t

p (1) = \frac{\sqrt{2}}{2} \int_{0}^{1} (\frac{t + \sqrt{2}}{{(\sqrt{2} t + 1)}^{2} + 1} - \frac{t - \sqrt{2}}{{(\sqrt{2} t - 1)}^{2} + 1}) d t

u = \sqrt{2} t + 1 \Rightarrow d u = \sqrt{2} d t

y = \sqrt{2} t - 1 \Rightarrow d y = \sqrt{2} d t

p (1) = \frac{\sqrt{2}}{4} (\int_{1}^{\sqrt{2} + 1} \frac{u + 1}{u^{2} + 1} d y - \int_{- 1}^{\sqrt{2} - 1} \frac{y - 1}{y^{2} + 1} d y)

= \frac{\sqrt{2}}{4} ({[\frac{1}{2} l n (u^{2} + 1) + {t a n}^{- 1} (u)]}_{1}^{\sqrt{2} + 1} - {[\frac{1}{2} l n (y^{2} + 1) - {t a n}^{- 1} (y)]}_{- 1}^{\sqrt{2} - 1})

= \frac{\sqrt{2}}{4} (\frac{1}{4} l n 2 + \frac{1}{2} l n (\sqrt{2} + 1) + {t a n}^{- 1} (\sqrt{2} + 1) - \frac{π}{4} - (\frac{1}{4} l n (2) + \frac{1}{2} l n (\sqrt{2} - 1) - {t a n}^{- 1} (\sqrt{2} - 1) - \frac{π}{4}))

= \frac{\sqrt{2}}{4} (l n (\sqrt{2} + 1) + {t a n}^{- 1} (\sqrt{2} + 1) + {t a n}^{- 1} (\sqrt{2} - 1))

= \frac{\sqrt{2}}{4} (l n (\sqrt{2} + 1) + \frac{3 π}{8} + \frac{π}{8})

\underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{4 n + 1} = p (1) = \frac{\sqrt{2}}{4} (l n (\sqrt{2} + 1) + \frac{π}{2})

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

t h a n k y o u s i r .