Closed form of power series

Question

First day on the site and this is an amazing place!

this is my question:

knowing that: $$1+\frac{x^2}{2}+\frac{x^4}{4}+\frac{x^6}{6}+\frac{x^8}{8}+\cdots $$

is a power series, how can I obtain its closed form?

Thank you.

Answer 1

Let $$f(x)=1+\frac{x^2}2+\frac{x^4}4+\frac{x^6}6+\cdots\ .$$ Then $$f'(x)=x+x^3+x^5+\cdots=\frac{x}{1-x^2}\ ,$$ so $$f(x)=\int\frac{x}{1-x^2}\,dx=-\frac12\ln(1-x^2)+C\ .$$ And since $f(0)=1$ we have $C=1$.

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Alternatively, if you know the standard series $$\ln(1+t)=t-\frac{t^2}2+\frac{t^3}3-\cdots\ ,$$ you can substitute $t=-x^2$ and then do some simple algebra.

Answer 2

David's answer is certainly the way to go. I will just expand a little about the result (for $|x|<1$)

$$ S =x + x^3 + x^5 + \cdots = \frac{x}{1-x^2} $$ Notice that in such case both series:

$$ S_1 =1 + x + x^2 +\cdots = \frac{1}{1-x} $$ $$ S_2 = 1 + x^2 + (x^2)^2 + (x^2)^3 + \cdots = \frac{1}{1-x^2} $$ converge absolutely, so we may write $S = S_1-S_2$ and substract termwise, one gets:

$$ S = \frac{1}{1-x} -\frac{1}{1-x^2} =\frac{x}{1-x^2} $$