find-the-rsdius-of-convergence-for-theserie-n-0-x-n-2n-1-and-calculate-its-sum-s-x-find-the-value-of-n-0-1-2-n-2n-1-

Question Number 26192 by abdo imad last updated on 22/Dec/17

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

\sum_{n = 0}^{\propto} \frac{x^{n}}{2 n + 1} a n d c a l c u l a t e i t s s u m s (x)

f i n d t h e v a l u e o f \sum_{n = 0}^{\propto} \frac{1}{2^{n} (2 n + 1)} .

Commented by prakash jain last updated on 22/Dec/17

r a t i o t e s t

\lim_{n \to \infty} ∣ \frac{x^{n + 1}}{2 n + 1} \cdot \frac{2 n}{x^{n}} ∣=∣ x ∣ \lim_{n \to \infty} \frac{2 n}{2 n + 1} =∣ x ∣

∣ x ∣< 1 \Rightarrow series converges

∣ x ∣> 1 \Rightarrow series diverges

____

x = 1

\sum_{n = 0}^{\infty} \frac{1}{2 n + 1}, we know it diverges

radius of convergence ∣ x ∣< 1

Commented by abdo imad last updated on 22/Dec/17

l e t p u t a_{n} (x) = \frac{x^{n}}{2 n + 1} p e r x n o t 0

/ \frac{a_{n + 1} (x)}{a_{n} (x)} / = / \frac{x^{n + 1}}{2 n + 3} {(\frac{x^{n}}{2 n + 1})}^{- 1} / = / \frac{2 n + 1}{2 n + 3} x / = \frac{2 n + 1}{2 n + 3} / x /

a n d {l i m}_{n - >\propto} / \frac{a_{n + 1} (x)}{a_{n (x)}} / = / x / \Rightarrow i f / x / < 1 t h e s e r i e i s c o n v e r g e n t

i f / x / > 1 t h e s e r i e d i v e r g e i f x = 1 a l s o t h e s e r i e d i v e r g e

i f x = - 1 \sum_{n ⩾ 0} \frac{{(- 1)}^{n}}{2 n + 1} i s c o n v e r g e n t b e c a u s e i t s a a l t e r n e s e r i e

e t S (x) = \sum_{n = 0}^{\propto} \frac{x^{n}}{2 n + 1} t h e s u m f o r / x / < 1

c a s e 1 0 < x < 1 S (x) = \sum_{n = 0}^{\propto} \frac{{(\sqrt{x})}^{2 n}}{2 n + 1} = \frac{1}{\sqrt{x}} \sum_{n = 0}^{\propto} \frac{{(\sqrt{x})}^{2 n + 1}}{2 n + 1}

= \frac{1}{\sqrt{x}} φ (\sqrt{x}) w i t h φ (t) = \sum_{n = 0}^{\propto} \frac{t^{2 n + 1}}{2 n + 1} a n d / t / < 1

\partial φ / \partial t = \sum_{n = 0}^{\propto} t^{2 n} = \frac{1}{1 - t^{2}} \Rightarrow φ (t) = \int \frac{d t}{1 - t^{2}} + λ

= \frac{1}{2} l n / \frac{1 + t}{1 - t} / + λ λ = φ (0) = 0 \Rightarrow S (x) = \frac{1}{2 \sqrt{x}} l n (\frac{1 + \sqrt{x}}{1 - \sqrt{x}})

c a s e 2 - 1 < x < 0 S (x) = \sum_{n = 0}^{\propto} \frac{{(- 1)}^{n} {(\sqrt{- x})}^{2 n}}{2 n + 1}

= \frac{1}{\sqrt{- x}} \sum_{n = 0}^{\propto} \frac{{(- 1)}^{n} {(\sqrt{- x})}^{2 n + 1}}{2 n + 1} = \frac{1}{\sqrt{- x}} ψ (\sqrt{- x}) w i t h

ψ (t) = \sum_{n = 0}^{\propto} \frac{{(- 1)}^{n} t^{2 n + 1}}{2 n + 1} a n d / t / < 1

\partial ψ / \partial t = \sum_{n = 0}^{\propto} {(- 1)}^{n} t^{2 n} = \sum_{n = 0}^{\propto} {(- t^{2})}^{n} = \frac{1}{1 + t^{2}}

\Rightarrow ψ (t) = \int \frac{d t}{1 + t^{2}} + κ = a r c t a n t + κ k = ψ (0) = 0 \Rightarrow

S (x) = \frac{a r t a n (\sqrt{- x})}{\sqrt{- x}} .

v a l u e o f S = \sum_{n = 0}^{\propto} \frac{1}{2^{n} (2 n + 1)} w e h a v e p r o v e d t h a t \sum_{n = 0}^{\propto} \frac{x^{n}}{2 n + 1} = \frac{1}{2 \sqrt{x}} l n (\frac{1 + \sqrt{x}}{1 - \sqrt{x}}) i f 0 x < 1

w e t a k e x = \frac{1}{2} \Rightarrow S = \frac{1}{2 \sqrt{\frac{1}{2}}} l n (\frac{1 + \sqrt{\frac{1}{2}}}{1 - \sqrt{\frac{1}{2}}})

= \frac{1}{\sqrt{2}} l n (\frac{1 + \sqrt{2}}{- 1 + \sqrt{2}})

Commented by prakash jain last updated on 22/Dec/17

Thanks