1.

Prove that `1+(1)/(sqrt2)+(1)/(sqrt3)+....+(1)/(sqrtn) ge sqrtn, AA n in N`

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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