This is from a book im reading.
$ 1-\frac{x^2}{3!}+ \frac{x^2}{5!} -\frac{x^2}{7!}+ \frac{x^2}{9!}-...$
$= [1-\frac{x^2}{\pi^2}]\ [1-\frac{x^2}{4\pi^2}]\ [1-\frac{x^2}{9\pi^2}]\ [1-\frac{x^2}{16\pi^2}]... $
$= 1-(\frac{1}{\pi^2}+ \frac{1}{4\pi^2}+ \frac{1}{9\pi^2}+ \frac{1}{16\pi^2}+...)x^2+(...)x^4-\ ... $
"once Euler had multiplied out the infinite product to get two infinite sums equaling each other, nothing would be more natural than to equate the like powers of x. Note that both series begin with 1 . Next comes the $x^2$ term in each series, and so their coefficients must be equal . That is,"
$$-\frac{1}{3!}= -(\frac{1}{\pi^2}+ \frac{1}{4\pi^2}+ \frac{1}{9\pi^2}+ \frac{1}{16\pi^2}\,+\,...) $$
What I dont understand is that where did this $-\frac{1}{3!}$ came from?? please help.
Thanks.
On one hand you have the series $1-\frac1{3!}x^2 + \cdots$. On the other you have the series $1 - \left(\frac1{\pi^2} + \frac1{4\pi^2} + \cdots\right)x^2 + \cdots$ These series are thought to be equal. The coefficient of $x^2$ on one side is $-\frac1{3!}$, while on the other side it's $\frac1{\pi^2} + \frac1{4\pi^2} + \cdots$. If the series as a whole are equal, then these two coefficients ought to be equal. So that's where the final equality comes from.