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