Watching a regular series

Algebra Level pending

If L = 1 − 1 2 + 1 3 − 1 4 + 1 5 − . . . L=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-... , what is a a , if 4 L = l n a 4L=ln \ a ?


The answer is 16.

Mikael Marcondes by Mikael Marcondes , 24, Brazil

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1 solution

Chew-Seong Cheong
May 24, 2015

Using the following Maclaurin series,

ln ⁡ ( 1 + x ) = x − 1 2 x 2 + 1 3 x 3 − 1 4 x 4 + 1 5 x 5 − . . . \ln{(1+x)} = x - \frac{1}{2}x^2 + \frac{1}{3}x^3 - \frac{1}{4}x^4 + \frac{1}{5}x^5 -...

Therefore,

L = 1 − 1 2 + 1 3 − 1 4 + 1 5 − . . . = ln ⁡ ( 1 + 1 ) = ln ⁡ 2 ⇒ 4 L = 4 ln ⁡ 2 = ln ⁡ ( 2 4 ) = ln ⁡ 16 = ln ⁡ a ⇒ a = 16 L = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} -... = \ln{(1+1)} = \ln{2} \\ \Rightarrow 4L = 4\ln{2} = \ln{\left(2^4\right)} = \ln{16} = \ln{a} \\ \Rightarrow a = \boxed{16}

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