Finding the sum of an infinite power series with formula summation.

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So I learned a formula which says that $\sum_{n=0}^{\infty} x^n= \frac{1}{1-x}$ which it can be used in fact to determine a sum of a power series.

I have this exercise to determine the sum: $$\sum_{n=1}^{\infty} \frac {x^{2n}}{2n}$$ for $|x| <1 $.

The answer given in the book is $-\frac 12 \cdot \ln(1-x^2)$.

But when I calculated, I got $2\ln|x| + \frac{1}{1-x}$.

Could you help me please, I would appreciate it very much. Thanks!

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There are 2 best solutions below

9
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Hint

$$y=\sum_{n=1}^{\infty} \frac {x^{2n}}{2n}\implies y'=\sum_{n=1}^{\infty} x^{2n-1}=\frac 1x \sum_{n=1}^{\infty} x^{2n}=\frac 1x \sum_{n=1}^{\infty} (x^2)^n$$

0
On

An alternative method:

$$\sum_{n=1}^{\infty}\frac{x^{2n}}{2n}=\sum_{n=1}^{\infty}\int x^{2n-1}dx=\int \sum_{n=1}^{\infty} x^{2n-1}dx=\int\frac{x}{1-x^2}dx=-\frac{1}{2}log(1-x^2)$$