Question

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

Answer 1

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.