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we know that $$\frac{\pi^2}{6}=\int_{0}^{\infty}\frac{t}{e^t-1}dt$$, we also know that $\frac{t}{e^t-1}$ is the generating function for the Bernoulli numbers i.e $ \frac{t}{e^t-1} =\sum_{n=1}^{\infty}\frac{B_nt^n}{n!}$. If we use this generating function in the above i.e $$\frac{\pi^2}{6}=\int_{0}^{\infty}\sum_{n=1}^{\infty}\frac{B_nt^n}{n!}dt$$ changing the order of integration $$\frac{\pi^2}{6}=\sum_{n=1}^{\infty}B_n\int_{0}^{\infty}\frac{t^n}{n!}dt$$ but this is not true. Where is the flaw.

user90533
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    $\frac{t}{e^t-1}$ is an analytic function but not an entire one. The radius of convergence in the origin is $2\pi$. – Jack D'Aurizio Mar 29 '16 at 17:31
  • but it has power series expansion, then the series is uniformly converges then, we can change the order of integration – user90533 Mar 29 '16 at 17:34
  • http://math.stackexchange.com/questions/391807/radius-of-convergence-of-the-bernoulli-polynomial-generating-function-power-seri – Winther Mar 29 '16 at 17:35
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    Restating Jack's comment: The series doesn't even converge for $t\ge2\pi$. Before you can interchange the integral and series, it's a requirement that the series makes sense for all values in the integral's domain. The power series does not converge uniformly to the function on the interval $[0,\infty)$, it doesn't even converge at all on $[2\pi,\infty)$. – anon Mar 29 '16 at 17:36
  • I got my mistake – user90533 Mar 29 '16 at 17:39

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