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!
$\endgroup$ 12 Answers
$\begingroup$Hint
$$y=\sum_{n=1}^{\infty} \frac {x^{2n}}{2n}\implies y'=\sum_{n=1}^{\infty} x^{2n-1}=\frac 1x \sum_{n=1}^{\infty} x^{2n}=\frac 1x \sum_{n=1}^{\infty} (x^2)^n$$
$\endgroup$ 10 $\begingroup$An alternative method:
$$\sum_{n=1}^{\infty}\frac{x^{2n}}{2n}=\sum_{n=1}^{\infty}\int x^{2n-1}dx=\int \sum_{n=1}^{\infty} x^{2n-1}dx=\int\frac{x}{1-x^2}dx=-\frac{1}{2}log(1-x^2)$$
$\endgroup$ 1
