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Use the series you know to show that:

ln3+(ln3)22!+(ln3)33!+…=2

Short Answer

Expert verified

It is shown that the expansion of seriesln(3)+ln(3)22!+ln(3)33!+ln(3)44!+…is2

Step by step solution

01

Maclaurin series and the given series.

Series isln(3)+ln(3)22!+ln(3)33!+ln(3)44!+…
The Maclaurin series ofis given below:

ex=1+x+x22!+x33!+x44!+…ex-1=x+x22!+x33!+x44!+…

02

Use the Maclaurin series.

Substituteln(3) forx in the Maclaurin series of ex-1=x+x22!+x33!+x44!+…as follows:

eln(3)-1=ln(3)+ln(3)22!+ln(3)33!+ln(3)44!+…

The value of eln(3)is given below:

eln(3)=3

Substitute 3foreln(3) in the equation eln(3)-1=ln(3)+ln(3)22!+ln(3)33!+ln(3)44!+…as follows:

3-1=ln(3)+ln(3)22!+ln(3)33!+ln(3)44!+…2=ln(3)+ln(3)22!+ln(3)33!+ln(3)44!+…

Thus, it is shown that the expansion of seriesln(3)+ln(3)22!+ln(3)33!+ln(3)44!+…is2

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