How do you find the sum of the series $1+ln2+(((ln2)^2)/(2!))+...+(((ln2)^2)/(n!))+...$ ?

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Answer 1 · 2015-02-26T10:24:32

The sum is $2$ .

Let's write the McLaurin Series of $y=2^x$ :

$2^x=2^0+(2^0ln2)/(1!)x+(2^0(ln2)^2)/(2!)x^2+...+(2^0(ln2)^n)/(n!)x^n+...$ .

If we put in this series $x=1$ , on the left the result is $2$ , on the right the result is the series of the exercise.

How do you find the sum of the series 1+ln2+(((ln2)^2)/(2!))+...+(((ln2)^2)/(n!))+...1+ln2+(((ln2)^2)/(2!))+...+(((ln2)^2)/(n!))+...?