Generating function of reciprocal odd numbers

Question

can you please help me? I'm trying to find which function would generate me all reciprocal odd numbers and I don't know how should I do it... I've tried all methods known to me :) $\sum_{n=0}^{\infty} \frac{1}{2n+1}x^{n}$

thanks

Answer 1

A few gentle hints: (1) do you know how to get the generating function for $\frac1n$ (i.e., $\sum_n\frac1nx^n$)? (2) Do you know how to get the generating function for $\frac{(-1)^n}n$? (You should be able to derive this from the answer to (1) with a little cleverness!) (3) What do you get if you take the sum of these two series? What happens if you take the difference of them?

Generically, this technique is usually known as series multisection, and it's another useful tool in the combinatorial toolbox.

Answer 2

Recall the series for $\ln$ \begin{eqnarray*} \ln(1+x) =\sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^n}{n} \\ -\ln(1-x) = \sum_{n=1}^{\infty} \frac{x^n}{n}. \\ \end{eqnarray*} Add these and all the "even" terms will cancel to give \begin{eqnarray*} \ln \left(\frac{1+x}{1-x}\right) = 2\sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}. \\ \end{eqnarray*} Now rescale the $x$ \begin{eqnarray*} \sum_{n=0}^{\infty} \frac{x^{n}}{2n+1} =\frac{1}{2\sqrt{x}} \ln \left(\frac{1+\sqrt{x}}{1-\sqrt{x}}\right). \\ \end{eqnarray*}