Power series summation Trying to find the sum of the following infinite series:
$$ \displaystyle\sum_{n=1}^{\infty}\frac{{(-1)}^{n-1}}{(2n-1)3^{n-1}}$$
Any ideas on how to find this sum?
 A: Note that for suitable $t$,
$$\frac{1}{1+t^2}=1-t^2+t^4-t^6+\cdots.$$
Integrate term by term from $0$ to $x$. We get
$$\arctan x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots.$$
Divide by $x$. We get
$$\frac{\arctan x}{x}=1-\frac{x^2}{3}+\frac{x^4}{5}-\frac{x^6}{7}+\cdots.$$
Finally, let $x=\frac{1}{\sqrt{3}}$.
A: Consider
$$f(x)=\sum_{n=1}^{\infty}\frac{(-x^2)^{n-1}}{2n-1}$$
Then
$$f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{2n-2}}{2n-1}$$
$$x\, f(x)=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}x^{2n-1}}{2n-1}$$
$$(x\, f(x))'=\sum_{n=1}^{\infty}{(-1)^{n-1}x^{2n-2}}=\sum_{n=1}^{\infty}{(-x^2)^{n-1}}=\sum_{n=0}^{\infty}{(-x^2)^n}=\frac{1}{1+x^2}$$
$$x\, f(x)=\int \frac{1}{1+x^2}={\tan}^{-1}x$$
$$f(x)=\frac{{\tan}^{-1}x}{x}$$
Your series is equal to
$$f\left(\frac{1}{\sqrt{3}}\right)$$
A: Hint: $\text{arctanh }x=\displaystyle\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1}$
A: $$
\frac{x^{n-1}}{2n-1} = \frac{y^{2n-2}}{2n-1} \quad \text{where }y=\sqrt x
$$
and
$$
\frac d {dy}\, \frac{y^{2n-1}}{2n-1} = y^{2(n-1)} = x^{n-1}.
$$
$$
\sum_{n=1}^\infty \frac{x^{n-1}}{2n-1} = \sum_{n=1}^\infty \frac{y^{2n-1}}{2n-1}.
$$
The derivative of this with respect to $y$ is
$$
\sum_{n=1}^\infty y^{2n-1} = \frac y {1-y^2} = \frac A {1-y} + \frac B {1+y}, 
$$
so find the antiderivative of that with respect to $y$ (you'll need to find $A$ and $B$) and after that, figure out what it has to do with the function of $x=y^2$.
A: You can consider the geometric sum $\sum_{n=0}^\infty \left(-\frac {x^2}3\right)^n$
This series is also $\sum_{k=2}^\infty \left(-\frac{x^2}3\right)^{k-1}=\sum_{k=1}^\infty \left(-\frac13\right)^{k-1} x^{2k-2}$ and an integration from 0 to x give
$\sum_{k=1}^\infty \left(-\frac13\right)^{k-1} \frac{x^{2k-1}}{2k-1}$. So you can calculate the sum of the series from the geometric sum. 
