Question Number 95145 by i jagooll last updated on 23/May/20
$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\: \\ $$$$\frac{\mathrm{1}}{\mathrm{14}}\:+\:\frac{\mathrm{1}}{\mathrm{35}}\:+\frac{\mathrm{1}}{\mathrm{65}}\:+\:\frac{\mathrm{1}}{\mathrm{104}}\:+\:...\:? \\ $$
Answered by bobhans last updated on 23/May/20
![S = Σ_(n = 1) ^∞ ((2/((3n+1)(3n+2)))) = (2/3) lim_(x→∞) Σ_(n = 1) ^x ((1/(3n+1)) −(1/(3n+4))) = (2/3) lim_(x→∞) [ (1/4)−(1/(3x+4)) ] = (2/3) × (1/4) = (1/6)](Q95148.png)
$$\mathrm{S}\:=\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\infty} {\sum}}\left(\frac{\mathrm{2}}{\left(\mathrm{3n}+\mathrm{1}\right)\left(\mathrm{3n}+\mathrm{2}\right)}\right)\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\underset{\mathrm{n}\:=\:\mathrm{1}} {\overset{\mathrm{x}} {\sum}}\left(\frac{\mathrm{1}}{\mathrm{3n}+\mathrm{1}}\:−\frac{\mathrm{1}}{\mathrm{3n}+\mathrm{4}}\right) \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{3}}\:\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:\left[\:\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{3x}+\mathrm{4}}\:\right]\:=\:\frac{\mathrm{2}}{\mathrm{3}}\:×\:\frac{\mathrm{1}}{\mathrm{4}}\:=\:\frac{\mathrm{1}}{\mathrm{6}} \\ $$
Commented by i jagooll last updated on 23/May/20
$$\mathrm{thank}\:\mathrm{you} \\ $$