Question Number 123108 by benjo_mathlover last updated on 23/Nov/20
Commented by benjo_mathlover last updated on 23/Nov/20
$$\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}+\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{3}}+\mathrm{3}\sqrt{\mathrm{2}}}+\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{5}}+\mathrm{5}\sqrt{\mathrm{4}}}+ \\ $$$$…\:+\:\frac{\mathrm{1}}{\mathrm{4012008}\sqrt{\mathrm{4012009}}+\mathrm{4012009}\sqrt{\mathrm{4012008}}}\:? \\ $$
Commented by liberty last updated on 23/Nov/20
$${we}\:{are}\:{asked}\:{to}\:{compute}\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{4012008}} {\sum}}\frac{\mathrm{1}}{{k}\sqrt{{k}+\mathrm{1}}+\left({k}+\mathrm{1}\right)\sqrt{{k}}}\: \\ $$$${consider}\:{the}\:{term}\:\frac{\mathrm{1}}{{k}\sqrt{{k}+\mathrm{1}}+\left({k}+\mathrm{1}\right)\sqrt{{k}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{k}}\:\sqrt{{k}+\mathrm{1}}\:\left(\sqrt{{k}}+\sqrt{{k}+\mathrm{1}}\right)} \\ $$$$=\:\frac{\sqrt{{k}+\mathrm{1}}−\sqrt{{k}}}{\:\sqrt{{k}}\:\sqrt{{k}+\mathrm{1}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{{k}}}\:−\:\frac{\mathrm{1}}{\:\sqrt{{k}+\mathrm{1}}} \\ $$$${so}\:{the}\:{equation}\:{becomes}\: \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{4012008}} {\sum}}\left(\frac{\mathrm{1}}{\:\sqrt{{k}}}−\frac{\mathrm{1}}{\:\sqrt{{k}+\mathrm{1}}}\right)\:=\:\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{4012009}}} \\ $$$$\:=\:\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{2003}^{\mathrm{2}} }}\:=\:\frac{\mathrm{2002}}{\mathrm{2003}}\:.\:\blacktriangle \\ $$$$ \\ $$