This can be proved by induction.

Sum to n terms is S[n], and we assume that S[n]>√n. Let S[n]=x+√n where x>0.

S[n+1]=S[n]+1/√(n+1)=S[n]+√(n+1)/(n+1).

Consider the expression S[n+1]-√(n+1)=S[n]+√(n+1)/(n+1)-√(n+1), which equals S[n]-n√(n+1)/(n+1) or S[n]-(n/(n+1))√(n+1).

This can be written S[n]-(√n)(1+1/n)^-½, because n√(n+1)/(n+1)=n/√(n+1)=n/√(n(1+1/n))=√n/√(1+1/n).

Substitute for S[n]: x+√n-(√n)(1+1/n)^-½ and expand the binomial to the first term:

x+√n-(√n)(1-1/(2n))=x+1/(2√n) which is a positive quantity. Further expansion of the binomial introduces smaller terms (positive and negative) as coefficients get smaller and negative powers of n create smaller fractions ((1+1/n)^-½=1-1/(2n)+3/(8n²)-5/(16n³)+35/(128n⁴)...).

Therefore (approximately) S[n+1]-√(n+1)=x+1/(2√n)>0 and S[n+1]>√(n+1) for n≥2.

But we need to prove a base case. S₂=1+1/√2=1.7071 which is greater than √2. So for higher values of n S[n]>√(n+1) QED.