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参考・概略です

(1)より

 1/(√1+√2)=√2-√1

 1/(√2+√3)=√3-√2

 1/(√3+√4)=√4-√3

 ・・・・・

 1/{√n+√(n+1)}=√(n+1)-√n

以上から

 S=(√2-√1)+(√3-√2)+(√4-√3)+…+(√(n+1)-√n)

 ●√2,√3,…,√nが消去され

 S=(-√1)+√(n+1)=√(n+1)-1

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