calculate-n-0-n-1-n-x-n-

Question Number 94335 by mathmax by abdo last updated on 18/May/20

c a l c u l a t e \sum_{n = 0}^{\infty} n^{{(- 1)}^{n}} x^{n}

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

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

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

we have \sum_{n = 0}^{\infty} x^{n} = \frac{1}{1 - x} \Rightarrow \sum_{n = 1}^{\infty} {nx}^{n - 1} = \frac{1}{{(1 - x)}^{2}} \Rightarrow

\sum_{n = 1}^{\infty} {nx}^{n} = \frac{x}{{(1 - x)}^{2}} \Rightarrow \sum_{n = 1}^{\infty} {nx}^{2 n} = \frac{x^{2}}{{(1 - x^{2})}^{2}} = h (x)

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

k (x) = \int_{0}^{x} \frac{dt}{1 - t^{2}} + k (k = 0) = \frac{1}{2} \int_{0}^{x} (\frac{1}{1 - t} + \frac{1}{1 + t}) dt

= \frac{1}{2} {[\ln ∣ \frac{1 + t}{1 - t} ∣]}_{0}^{x} = \frac{1}{2} \ln ∣ \frac{1 + x}{1 - x} ∣ \Rightarrow s (x) = \frac{{2 x}^{2}}{{(1 - x^{2})}^{2}} + \frac{1}{2} \ln ∣ \frac{1 + x}{1 - x} ∣

with ∣ x ∣ < 1

Answered by maths mind last updated on 18/May/20

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

s = \sum_{n ⩾ 1} 2 {n x}^{2 n} - x \sum_{n ⩾ 0} 2 {n x}^{2 n} - x \sum_{n ⩾ 0} x^{2 n}

\sum_{n ⩾ 1} {n x}^{n - 1} = \frac{1}{{(1 - x)}^{2}}

\Rightarrow Σ 2 {n x}^{2 n} = \frac{2 x^{2}}{{(1 - x^{2})}^{2}}

\Rightarrow s = \frac{2 x^{2}}{{(1 - x^{2})}^{2}} - \frac{2 x^{3}}{{(1 - x^{2})}^{2}} - \frac{x}{1 - x^{2}}

s = \frac{- x^{3} + 2 x^{2} - x}{{(1 - x^{2})}^{2}}

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

thanks sir .