Question Number 95473 by aurpeyz last updated on 25/May/20
Answered by EmericGent last updated on 25/May/20
$${regroup}\:{every}\:{two}\:{terms} \\ $$$$\mathrm{1}-\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${and}\:{then}\:\forall{n}\:\in\:\mathbb{N}\ast \\ $$$$\frac{\mathrm{1}}{{n}}\:−\:\frac{\mathrm{1}}{{n}+\mathrm{1}}\:=\:\frac{{n}+\mathrm{1}−{n}}{{n}\left({n}+\mathrm{1}\right)}\:=\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)} \\ $$$${since}\:{n}\left({n}+\mathrm{1}\right)\:{is}\:{a}\:{non}\:{zero}\:{positive}\: \\ $$$${integer},\:\frac{\mathrm{1}}{{n}\left({n}+\mathrm{1}\right)}\:{is}\:{greater}\:{than}\:\mathrm{0} \\ $$$${so}\:{your}\:{sum}\:{is}\:{just}\:{a}\:{sum}\:{of}\:{positive} \\ $$$${terms},\:{so}\:{it}'{s}\:{also}\:{positive} \\ $$
Commented by aurpeyz last updated on 06/Jun/20
$${Please}\:{explain}.\:{i}\:{do}\:{not}\:{seem}\:{to}\:{understand}.\:{pla} \\ $$