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k-0-1-4k-1-1-4k-3-




Question Number 212359 by MrGaster last updated on 11/Oct/24
           Σ_(k=0) ^∞ ((1/( (√(4k+1))))−(1/( (√(4k+3)))))
$$ \\ $$$$\:\:\:\:\:\:\:\:\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{k}+\mathrm{1}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{4}{k}+\mathrm{3}}}\right) \\ $$
Answered by MathematicalUser2357 last updated on 13/Oct/24
0.622791226505... ...
$$\mathrm{0}.\mathrm{622791226505}…\:… \\ $$
Commented by Ghisom last updated on 14/Oct/24
exact solution possible and required
$$\mathrm{exact}\:\mathrm{solution}\:\mathrm{possible}\:\mathrm{and}\:\mathrm{required} \\ $$
Answered by Berbere last updated on 15/Oct/24
=Σ_(k≥0) (1/2)((1/( (√(k+(1/4)))))−(1/( (√(k+(3/4))))))=(1/2)(ζ((1/2),(1/4))−ζ((1/2),(3/4)))  ζ(s,z)=Σ_(n≥0) (1/((n+z)^s ))   zeta hurwitz function
$$=\underset{{k}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\:\sqrt{{k}+\frac{\mathrm{1}}{\mathrm{4}}}}−\frac{\mathrm{1}}{\:\sqrt{{k}+\frac{\mathrm{3}}{\mathrm{4}}}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\zeta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{1}}{\mathrm{4}}\right)−\zeta\left(\frac{\mathrm{1}}{\mathrm{2}},\frac{\mathrm{3}}{\mathrm{4}}\right)\right) \\ $$$$\zeta\left({s},{z}\right)=\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{1}}{\left({n}+{z}\right)^{{s}} }\:\:\:{zeta}\:{hurwitz}\:{function} \\ $$

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