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 $f$.

  • $\begingroup$ 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. $\endgroup$ – Tunococ Sep 4 '13 at 23:32
  • $\begingroup$ Yep but I don't know how to expand each side as a power series $\endgroup$ – Lucas Zanella Sep 4 '13 at 23:33
  • $\begingroup$ I am not sure what you mean by "expand". $\endgroup$ – Tunococ Sep 4 '13 at 23:42
  • $\begingroup$ You're saying that both my left and my right sides of the equation are an infinite expansion of some function, right? $\endgroup$ – Lucas Zanella Sep 4 '13 at 23:54
  • 1
    $\begingroup$ 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. $\endgroup$ – Tunococ Sep 4 '13 at 23:58

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