Question

Cho \(S=\frac{1}{\sqrt{1.1998}}+\frac{1}{\sqrt{2.1997}}+..+\frac{1}{\sqrt{k\left(k.1998-k+1\right)}}+\frac{1}{\sqrt{1998-1}}\)

Hãy so sánh S và \(2\frac{1998}{1999}\)

Answer 1

\(2\frac{1998}{1999}\)là hỗn số hay \(2.\frac{1998}{1999}\)hả bạn?

Answer 2

Là \(2.\frac{1998}{1999}\)

Answer 3

ok bạn đợi mình tí nhé :>

Answer 4

\(S=\frac{1}{\sqrt{1.1998}}+\frac{1}{\sqrt{2.1997}}+...+\frac{1}{\sqrt{k.\left(1998-k+1\right)}}+\frac{1}{\sqrt{1998.1}}\)

áp dụng bđt Cô si \(\frac{a+b}{2}>\sqrt{ab}\Rightarrow\frac{2}{a+b}< \frac{1}{\sqrt{ab}}\)(\(a\ne b\); \(a;b>0\))

\(\frac{1}{\sqrt{1.1998}}>\frac{2}{1+1998}=\frac{2}{1999}\)

\(\frac{1}{\sqrt{2.1997}}>\frac{2}{2+1997}=\frac{2}{1999}\)

...

\(\frac{1}{\sqrt{1998.1}}>\frac{2}{1998+1}=\frac{2}{1999}\)

cộng vế theo vế ta được

S > \(1998.\frac{2}{1999}\)

hay \(S>2.\frac{1998}{1999}\)