# Infinite sum and equality between coefficients of the same index

I have two infinite sums that forms an equality:

$$\sum_{n=1}^{\infty} \left(\zeta(2n)\frac{x^{2n}}{\pi^{2n}}\right) = \sum_{n=1}^\infty \left(\frac{B_{2n}}{(2n)!}\left(-\frac{1}{2}\right)(2ix)^{2n}\right)$$

I know that the entire sum of the left side is equals to the entire sum of the right side. But how do I know that if I expand the left side, the coefficient of index 1 is gonna be equals to the coefficient of index 1 in the right side? In other words, proof that:

$$\zeta(2n)\frac{x^{2n}}{\pi^{2n}} = \frac{B_{2n}}{(2n)!}\left(-\frac{1}{2}\right)(2ix)^{2n}$$

If a smooth function $f$ can be expressed as a power series around $0$: $$f(x) = \sum_{n=0}^\infty a_n x^n,$$ then $a_n = \frac{1}{n!}f^{(n)}(0)$, where $f^{(n)}$ is the $n$-th derivative of $f$.
• I am saying by taking the $n$-th derivative, dividing by $n!$, and substituting $x = 0$, you can "extract" the $n$-th coefficient of a power series. That is how you justify that coefficients of two equal power series will have to be equal. – Tunococ Sep 4 '13 at 23:32
• If you assume the power series on both sides are convergent, then both sides are analytic functions near $0$. You can find, for each $n$, the $n$-th derivatives of both functions, which are also functions analytic near $0$. If you substitute $x = 0$ in the derivatives, you will get the $n$-th coefficients of the original power series (multiplied by $n!$) on both sides. In short, if the original power series are equal, their derivatives of all orders must be equal, and so are their coefficients. – Tunococ Sep 4 '13 at 23:58