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Prove that `1+(1)/(sqrt2)+(1)/(sqrt3)+....+(1)/(sqrtn) ge sqrtn, AA n in N` |
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Answer» For n=1, so LHS=RHS ….(i) Assume the result for n=k `i.e. 1+(1)/(sqrt2)+(1)/(sqrt3)+.....+(1)/(sqrtk) ge sqrtk ....(ii)` For n=k+1 LHS=`1+(1)/(sqrt2)+..+(1)/(sqrtk)+(1)/(sqrt(k+1))` `ge sqrtk+(1)/(sqrt(k+1))" "["using (ii)"]` `gt sqrtk+(1)/(sqrt(k+1)+sqrtk)"Note"` `=sqrtk+sqrt((k+1))-sqrtk=sqrt((k+1))` i.e. the result is true for n=k+1 Hence, by induction, the result is true `AA n in N`. |
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