黎曼猜测的平凡零点不存在

伯努利数html ( 1 ) ∑ x = 1 n x = 1 + 2 + 3 + ⋯ + n = n ( n + 1 ) 2 (1)\sum_{x=1}^{n}{x}=1+2+3+\cdots+n=\frac{n(n+1)}{2} (1)∑x=1n​x=1+2+3+⋯+n=2n(n+1)​ ∫ − 1 0 x ( x + 1 ) 2 d x = − 1 12 \int_{-1}^{0} \frac

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