find-the-sum-of-x-2x-2-3x-3-4x-4-5x-5-6x-6-where-x-lt-1-

Question Number 145996 by iloveisrael last updated on 10/Jul/21

find the sum of

x - {2 x}^{2} + {3 x}^{3} - {4 x}^{4} + {5 x}^{5} - {6 x}^{6} + \dots

where ∣ x ∣ < 1

Answered by mathmax by abdo last updated on 10/Jul/21

S (x) = \sum_{n = 1}^{\infty} n {(- 1)}^{n - 1} x^{n} \Rightarrow S (x) = - \sum_{n = 1}^{\infty} n {(- x)}^{n}

we have \sum_{n = 0}^{\infty} {(- x)}^{n} = \frac{1}{1 + x} by derivation we get

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

S (x) = \frac{x}{{(1 + x)}^{2}}

Answered by mr W last updated on 10/Jul/21

S = x - {2 x}^{2} + {3 x}^{3} - {4 x}^{4} + {5 x}^{5} - {6 x}^{6} + \dots

x S = x^{2} - {2 x}^{3} + {3 x}^{4} - {4 x}^{5} + {5 x}^{6} + \dots

(1 + x) S = x - x^{2} + x^{3} - x^{4} + x^{5} - x^{6} + \dots

(1 + x) S = \frac{x}{1 + x}

\Rightarrow S = \frac{x}{{(1 + x)}^{2}}

Answered by qaz last updated on 10/Jul/21

\underset{n = 1}{\sum^{\infty}} {(- 1)}^{n - 1} {nx}^{n}

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

= x (yD + 1) ∣_{y = - x} \underset{n = 0}{\sum^{\infty}} y^{n}

= x (yD + 1) ∣_{y = - x} \frac{1}{1 - y}

= x {[\frac{y}{{(1 - y)}^{2}} + \frac{1}{1 - y}]}_{y = - x}

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

Commented by qaz last updated on 10/Jul/21

D = \frac{d}{dx}

Dx - xD = 1

xDxD \cdot \dots \cdot xD = {(xD)}^{n} = {(Dx - 1)}^{n} = \underset{k = 0}{\sum^{n}} (\begin{matrix} n \\ k \end{matrix}) {(Dx)}^{k} {(- 1)}^{n - k}

D^{n} x - {xD}^{n} = {nD}^{n - 1}

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