Almost a geometric series, but not quite

Algebra Level 4

Let $f : (-1, 1) \to \mathbb{R}$ be defined as $\displaystyle f(x) = \sum_{n=1}^{\infty} nx^n$ .

Find the value of $\dfrac{1}{f \left(\dfrac{1}{5} \right) + f \left(-\dfrac{1}{3} \right)}$ .

The answer is 8.

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U Z
Jan 14, 2015

$g(x) = 1 + x + x^2 + x^3 + ...... = \dfrac{1}{1-x}$

$g'(x) = 1 + 2x + 3x^2 + ......... = \dfrac{1}{(1-x)^2}$

$xg'(x) = x + 2x^2 + 3x^3 + 4x^4 + .... = \dfrac{x}{(1-x)^2}$

$\displaystyle f(x) = \sum_{n=1}^{\infty} nx^n = xg'(x) = \dfrac{x}{(1-x)^2}$

$f\left (\dfrac{1}{5}\right ) = \dfrac{\dfrac{1}{5}}{\left (\dfrac{4}{5}\right )^2} = \dfrac{ 5}{16}$

$f\left (\dfrac{-1}{3}\right ) = -\dfrac{\dfrac{1}{3}}{\left (\dfrac{4}{3}\right )^2}= \dfrac{-3}{16}$

$f\left (\dfrac{1}{5}\right ) + f\left (\dfrac{-1}{3}\right ) = \dfrac{2}{16} = \dfrac{1}{8}$

Or let $S=x+2x^2+3x^3+\cdots$

We see $xS=x^2+2x^3+\cdots$

Subtracting, gives $(1-x)S=x+x^2+x^3+\cdots =\dfrac{x}{1-x}$

Thus, $S=\dfrac{x}{(1-x)^2}$

Daniel Liu - 6 years, 4 months ago

Nice. I think you meant to put the "+ ....." after $x + x^{2} + x^{3}$ , though.

Brian Charlesworth - 6 years, 4 months ago

Hey! I edited LaTeX a bit. You may use $\text{\left}$ and $\text{\right}$ before parentheses to get brackets of correct size. See your solution for reference.

Pranjal Jain - 6 years, 4 months ago

Almost a geometric series, but not quite

The answer is 8.

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