Can somebody prove this infinite series? Transformation of the Leibniz formula for $\pi$ results in the infinite series:
$$
\frac 1 {1 \times 3} + \frac 1 {5 \times 7} + \frac 1 {9 \times 11} + \frac 1 {13 \times 15} +\cdots =  \frac \pi 8
$$
If you recombine the numbers in the denominator you get e.g. the following series:
$$
\frac 1 {1 \times 5} + \frac 1 {3 \times 7} + \frac 1 {9 \times 13} + \frac 1 {11 \times 15} + \cdots
$$
which seems to approach $\dfrac{\pi / 8}{\sqrt 2}$.
Can somebody prove this?
 A: The same type of method as used to prove that $\pi/4 = 1 - 1/3 + 1/5 - 1/7 + \ldots$ can be used here, as soon as you write $\frac1{(2n + 1)(2n + 5)} = \frac1{4}(\frac1{2n + 1} - \frac1{2n + 5})$. 
So your series is 
$$\frac1{4}\left(1 + \frac1{3} - \frac1{5} - \frac1{7} + \ldots\right)$$ 
where the expression inside parentheses is $f(1)$ where $f(x) = x + x^3/3 - x^5/5 - x^{7}/7 + \ldots$. We calculate 
$$\begin{array}{ccl}
f'(x) & = & 1 + x^2 - x^4 - x^6 + \ldots \\ 
 & = & (1 + x^2)(1 - x^4 + x^8 - x^{12} + \ldots) \\ 
 & = & \frac{1 + x^2}{1 + x^4}
\end{array}
$$ 
and so, since $f(0) = 0$, we have 
$$f(1) = \int_0^1 \frac{1 + x^2}{1 + x^4} dx$$
Mathematica can give you the answer to this pretty quickly, but if you want to do this by hand, you can. We can factor $1 + x^4 = (1 - \sqrt{2} x + x^2)(1 + \sqrt{2} x + x^2)$; use a partial fraction decomposition to obtain 
$$\frac{1 + x^2}{1 + x^4} = \frac{1/2}{1 - \sqrt{2} x + x^2} + \frac{1/2}{1 + \sqrt{2} x + x^2}$$
which after a little work integrates to 
$$\frac{\sqrt{2}}{2} \left(\arctan(\frac{x - \sqrt{2}/2}{\sqrt{2}/2}) + \arctan(\frac{x + \sqrt{2}/2}{\sqrt{2}/2})\right)$$ 
and now plug in $1$ and $0$ and subtract, etc. to obtain 
$$f(1) = \frac{\sqrt{2}}{2}\left(\arctan(\sqrt{2} - 1) + \arctan(\sqrt{2} + 1)\right)$$ 
The expressions inside the arctangents are reciprocals of each other, so the arctangent angles are complementary: they sum to $\pi/2$. Hence $f(1) = \sqrt{2}\pi/4$, and your answer is $f(1)/4 = \sqrt{2}\pi/16$, as advertised. 
