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Calculus Level 3

1 2 x 2 + 2 3 x 3 + 3 4 x 4 + ⋯ \large \dfrac{1}{2} x^2 + \dfrac{2}{3} x^3 + \dfrac{3}{4} x^4 + \cdots

If 0 < x < 1 0<x<1 , then find the closed form of the infinite series above.

x 1 − x + ln ⁡ ( 1 − x ) \frac{x}{1-x} +\ln(1-x) ln ⁡ 1 + x 1 − x \ln\frac{1+x}{1-x} x 1 − x + ln ⁡ ( 1 + x ) \frac{x}{1-x} + \ln(1+x) 1 1 − x + ln ⁡ ( 1 − x ) \frac{1}{1-x} + \ln(1-x)

Akash Patalwanshi by akash patalwanshi , 28, India

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1 solution

Akash Patalwanshi
Apr 30, 2016

We know, x 1 − x = x + x 2 + x 3 + . . . \frac{x}{1-x} = x + x^2 + x^3 +... for 0 < x < 1 0 <x<1

and

log ⁡ ( 1 − x ) = − ( x + x 2 2 + x 3 3 + . . . ) \log(1-x) = -(x + \frac{x^2}{2} + \frac{x^3}{3} +...)
for 0 < x < 1 0<x<1

So, x 1 − x + log ⁡ ( 1 − x ) = x − x + x 2 − x 2 2 + x 3 − x 3 3 + . . . \frac{x}{1-x} + \log(1-x) = x - x + x^2 - \frac{x^2}{2} + x^3 - \frac{x^3}{3} +... for 0 < x < 1 0<x<1

→ x 1 − x + log ⁡ ( 1 − x ) = 1 2 x 2 + 2 3 x 3 + 3 4 x 4 + . . . . →\boxed{\frac{x}{1-x} + \log(1-x)} = \frac{1}{2}x^2 + \frac{2}{3}x^3 + \frac{3}{4}x^4 +.... for 0 < x < 1 0<x<1

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