Von Staudt-Clausen theorem

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Problem 8.182 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 6 "Power Series with Integral Coefficients in the Sense of Hurwitz"

The Bernoulli numbers BnB_n are defined as the coefficients in the power series expansion

zez−1=1−z2+∑n=1∞(−1)n−1Bn(2n)!z2n \frac{z}{e^z-1} = 1-\frac z2+\sum _{n=1}^\infty \frac{(-1)^{n-1}B_n}{(2n)!}z^{2n}

(Problem 154 of Part I). We have

B1=16=1−12−13,B2=130=−1+12+13+15, B_1=\frac16 = 1-\frac12-\frac13 , \qquad B_2=\frac{1}{30} = -1+\frac12+\frac13+\frac15 , B3=142=1−12−13−17,B4=130=−1+12+13+15, B_3=\frac{1}{42} = 1-\frac12-\frac13-\frac17 , \qquad B_4=\frac{1}{30} = -1+\frac12+\frac13+\frac15 , B5=566=1−12−13−111,B6=6912730=−1+12+13+15+17+113, B_5=\frac{5}{66} = 1-\frac12-\frac13-\frac{1}{11} , \qquad B_6=\frac{691}{2730} = -1+\frac12+\frac13+\frac15+\frac17+\frac{1}{13} , B7=76=2−12−13,B8=3617510=6+12+13+15+117. B_7=\frac76 = 2-\frac12-\frac13 , \qquad B_8=\frac{3617}{510} = 6+\frac12+\frac13+\frac15+\frac{1}{17} .

In general we have

Bn=Gn+(−1)n(12+13+1α+1β+1γ+⋯ ), B_n = G_n+(-1)^n\left(\frac12+\frac13+\frac1\alpha +\frac1\beta +\frac1\gamma +\cdots \right) ,

where GnG_n is an integer and 2,3,α,β,γ,…2,3,\alpha ,\beta ,\gamma ,\ldots denote those prime numbers that exceed by one the divisors of 2n2n. [Expand first zz and then z/(ez−1)z/(e^z-1) in ascending powers of (ez−1)(e^z-1) and apply 179--181.]

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Problem 8.262 from Pólya G., Szegő G. "Problems and Theorems in Analysis" [PóSz98]More problems in Section 2 "Miscellaneous Problems"

Denote by qnq_n the denominator of the nnth Bernoulli number BnB_n [182]. Then we have

lim⁡n→∞q1q2q3⋯qn(2n)!n=12. \lim _{n\to \infty } \sqrt[n]{\frac{q_1q_2q_3\cdots q_n}{(2n)!}} = \frac12 .
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References

[PóSz98] Pólya G., Szegő G. Problems and Theorems in Analysis. Springer, 1998.