The ln ⁡ 2 \ln 2 series

Calculus Level 2

1 + ln ⁡ 2 + ( ln ⁡ 2 ) 2 2 ! + ( ln ⁡ 2 ) 3 3 ! + ⋯ 1+\ln2+\frac{(\ln 2)^2}{2!}+\frac{(\ln 2)^3}{3!}+\cdots

The series above has a closed form. Find the value of this closed form.

Give your answer to 2 decimal places.


The answer is 2.00.

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Dmitri Ivanovich by Dmitri Ivanovich , 19, Japan

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2 solutions

Sam Bealing
May 24, 2016

e x = ∑ i = 0 ∞ x i i ! = 1 + x + x 2 2 ! + x 3 3 ! + ⋯ e^x=\sum_{i=0}^{\infty} \dfrac{x^i}{i!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+ \cdots

1 + ln ⁡ 2 + ( ln ⁡ 2 ) 2 2 ! + ⋯ = e ( ln ⁡ 2 ) = 2 1+\ln{2}+\dfrac{(\ln{2})^2}{2!}+ \cdots=e^{(\ln{2})}=\boxed{2}

7 Helpful 0 Interesting 1 Brilliant 0 Confused

Moderator note:

Simple standard approach.

e x = 1 + x + x 2 + . . . . . e^x=1+x+x^2+.....

Putting x = l n 2 x=ln2 we get e l n 2 e^{ln2} which is 2 \boxed{2}

0 Helpful 0 Interesting 0 Brilliant 0 Confused

You missed out dividing by the factorials in your e x e^x series.

Sam Bealing - 5 years ago

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