Question

20%20%20%5Cfrac%7B%20%7B%5Cpi%20%20%7D%5E%7B2n%20%2B%201%7D%20%7D%7B%282n%20-%201%29%21%282n%20%2B%201%29%7D%20%20%5C%5C%20" id="TexFormula1" title=" \rm\sum_{n = 1}^ \infty ( - 1 {)}^{n - 1} \frac{ {\pi }^{2n + 1} }{(2n - 1)!(2n + 1)} \\ " alt=" \rm\sum_{n = 1}^ \infty ( - 1 {)}^{n - 1} \frac{ {\pi }^{2n + 1} }{(2n - 1)!(2n + 1)} \\ " align="absmiddle" class="latex-formula"> ​

Answer 1

Let

$\displaystyle f(x) = \sum_{n=1}^\infty (-1)^{n-1} \frac{x^{2n-1}}{(2n-1)! (2n+1)}$

The exponent is indeed $2n-1$ - not a typo!

Take the antiderivative of $f$, denoted by $F$. This recovers a factor of $2n$ in the denominator, which lets us condense it to a single factorial.

$\displaystyle F(x) = \int f(x) \, dx = C + \sum_{n=1}^\infty (-1)^{n-1} \frac{x^{2n}}{(2n+1)!}$

Recall the series expansion of sine,

$\displaystyle \sin(x) = \sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!}$

Then with a little algebraic manipulation, we get

$\displaystyle F(x) = \int f(x) \, dx = C + 1 - \frac{\sin(x)}x$

Differentiate to recover $f$.

$f(x) = \dfrac{\sin(x) - x\cos(x)}{x^2}$

Finally, $f(\pi) = \frac1\pi$, so our sum is

$\displaystyle \pi^2 f(\pi) = \sum_{n=1}^\infty (-1)^{n-1} \frac{\pi^{2n+1}}{(2n-1)! (2n+1)} = \boxed{\pi}$