k-1-1-k-1-1-2k-1-2n-1-pi-2n-1-2-2n-2-2n-E-2m-

Question Number 222599 by MrGaster last updated on 01/Jul/25

Σ_(k=1) ^∞ (−1)^(k+1) (1/((2k−1)^(2n+1) ))=(π^(2n+1) /(2^(2n+2) (2n)!))∣E_(2m) ∣

$$\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\pi^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}\right)!}\mid{E}_{\mathrm{2}{m}} \mid \\ $$

Commented by MathematicalUser2357 last updated on 01/Jul/25

Prove Σ_(k=1) ^∞ (−1)^(k+1) (1/((2k−1)^(2n+1) ))=(π^(2n+1) /(2^(2n+2) (2n)!))∣E_(2m) ∣, so?

$$\mathrm{Prove}\:\underset{{k}=\mathrm{1}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{k}+\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{2}{k}−\mathrm{1}\right)^{\mathrm{2}{n}+\mathrm{1}} }=\frac{\pi^{\mathrm{2}{n}+\mathrm{1}} }{\mathrm{2}^{\mathrm{2}{n}+\mathrm{2}} \left(\mathrm{2}{n}\right)!}\mid{E}_{\mathrm{2}{m}} \mid,\:\mathrm{so}? \\ $$

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