1-calculate-n-0-x-n-4n-2-1-with-x-lt-1-2-find-the-value-of-n-0-1-4n-2-1-and-n-0-1-n-4n-2-1-

Question Number 94331 by mathmax by abdo last updated on 18/May/20

1) c a l c u l a t e \sum_{n = 0}^{\infty} \frac{x^{n}}{4 n^{2} - 1} w i t h ∣ x ∣< 1

2) f i n d t h e v a l u e o f \sum_{n = 0}^{\infty} \frac{1}{4 n^{2} - 1} a n d \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{4 n^{2} - 1}

Answered by mathmax by abdo last updated on 19/May/20

1) let f (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{{4 n}^{2} - 1} \Rightarrow f (x) = \sum_{n = 0}^{\infty} \frac{x^{n}}{(2 n - 1) (2 n + 1)}

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

case (1) 0 < x < 1 \Rightarrow \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 1} = \frac{1}{\sqrt{x}} \sum_{n = 0}^{\infty} \frac{{(\sqrt{x})}^{2 n + 1}}{2 n + 1} = \frac{1}{\sqrt{x}} φ (\sqrt{x)}

with φ (t) = \sum_{n = 0}^{\infty} \frac{t^{2 n + 1}}{2 n + 1} we have φ^{^{'}} (t) = \sum_{n = 0}^{\infty} t^{2 n} = \frac{1}{1 - t^{2}} \Rightarrow

φ (t) = \int \frac{dt}{1 - t^{2}} + c = \frac{1}{2} \int (\frac{1}{1 - t} + \frac{1}{1 + t}) dt + c = \frac{1}{2} \ln ∣ \frac{1 + t}{1 - t} ∣ + c

φ (0) = 0 = c \Rightarrow φ (t) = \frac{1}{2} \ln ∣ \frac{1 + t}{1 - t} ∣ \Rightarrow \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 1} = \frac{1}{2 \sqrt{x}} \ln ∣ \frac{1 + \sqrt{x}}{1 - \sqrt{x}} ∣

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

= x \times \frac{1}{2 \sqrt{x}} \ln ∣ \frac{1 + \sqrt{x}}{1 - \sqrt{x}} ∣ = \frac{\sqrt{x}}{2} \ln ∣ \frac{1 + \sqrt{x}}{1 - \sqrt{x}} ∣ \Rightarrow

f (x) = - \frac{1}{2} + \frac{\sqrt{x}}{4} \ln ∣ \frac{1 + \sqrt{x}}{1 - \sqrt{x}} ∣ - \frac{1}{4 \sqrt{x}} \ln ∣ \frac{1 + \sqrt{x}}{1 - \sqrt{x}} ∣

case (2) if - 1 < x < 0 we have f (x) = \frac{1}{2} \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} {(- x)}^{n}}{2 n - 1}

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

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

= \frac{1}{\sqrt{- x}} w (\sqrt{- x}) with w (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{2 n + 1} t^{2 n + 1} \Rightarrow

w^{^{'}} (x) = \sum_{n = 0}^{\infty} {(- 1)}^{n} t^{2 n} = \frac{1}{1 + t^{2}} \Rightarrow w (x) = arctant + c

w (0) = 0 = c \Rightarrow w (x) = \arctan (t) \Rightarrow \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n + 1} = \frac{1}{\sqrt{- x}} \arctan (\sqrt{- x})

wf follow the same way to find \sum_{n = 0}^{\infty} \frac{x^{n}}{2 n - 1} \dots .

Answered by mathmax by abdo last updated on 19/May/20

2) let S_{n} = \sum_{k = 0}^{n} \frac{1}{{4 k}^{2} - 1} \Rightarrow S_{n} = - 1 + \sum_{k = 1}^{n} \frac{1}{(2 k - 1) (2 k + 1)}

= - 1 + \frac{1}{2} \sum_{k = 1}^{n} (\frac{1}{2 k - 1} - \frac{1}{2 k + 1}) = - 1 + \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{2 k - 1} - \frac{1}{2} \sum_{k = 1}^{n} \frac{1}{2 k + 1}

\sum_{k = 1}^{n} \frac{1}{2 k - 1} = 1 + \frac{1}{3} + \dots . . + \frac{1}{2 n - 1} = 1 + \frac{1}{2} + \frac{1}{3} + \dots . + \frac{1}{2 n - 1} + \frac{1}{2 n}

- \frac{1}{2} - \frac{1}{4} - \dots - \frac{1}{2 n} = H_{2 n} - \frac{1}{2} H_{n}

\sum_{k = 1}^{n} \frac{1}{2 k + 1} =_{k = p - 1} \sum_{p = 2}^{n + 1} \frac{1}{2 p - 1} = H_{2 n + 2} - \frac{1}{2} H_{n + 1} - 1 \Rightarrow

S_{n} = - 1 + \frac{1}{2} H_{2 n} - \frac{1}{4} H_{n} - \frac{1}{2} H_{2 n + 2} + \frac{1}{4} H_{n + 1} + \frac{1}{2}

= - \frac{1}{2} + \frac{1}{2} (H_{2 n} - H_{2 n + 2}) + \frac{1}{4} (H_{n + 1} - H_{n}) but

\lim_{n ⌣ + \infty} H_{2 n} - H_{2 n + 2} = 0 = \lim_{n \to \infty} H_{n + 1} - H_{n} \Rightarrow

\lim_{n \to + \infty} S_{n} = - \frac{1}{2}