Áp dụng \(\frac{1}{\sqrt{a.b}}>\frac{2}{a+b}\) , ta có :
\(S=\frac{1}{\sqrt{1.1998}}+\frac{1}{\sqrt{2.1997}}+...+\frac{1}{\sqrt{k\left(1998-k+1\right)}}+...+\frac{1}{\sqrt{1998.1}}>\)
\(>\frac{2}{1+1998}+\frac{2}{2+1997}+...+\frac{2}{k+1998-k+1}+...+\frac{2}{1998+1}=\)
\(=\frac{2.1998}{1999}\)
Vậy \(S>\frac{2.1998}{1999}\)
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 \(2\frac{1998}{1999}\)
Cho S=\(\frac{1}{\sqrt{1.1998}}+\frac{1}{\sqrt{2.1997}}+......+\frac{1}{\sqrt{k\left(1998-k+1\right)}}+...+\frac{1}{\sqrt{1998.1}}\) hãy so sánh S và \(2\frac{1998}{1999}\)
Cho: \(S=\frac{1}{\sqrt{1.1998}}+\frac{1}{\sqrt{2.1997}}+...+\frac{1}{\sqrt{k.\left(1998-k+1\right)}}+...+\frac{1}{\sqrt{1998-1}}\)
Hãy so sánh: \(S\) và \(2.\frac{1998}{1999}\)
Tính
1) \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{1999\sqrt{1998}+1998\sqrt{1999}}\)
2) \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{1998}+\sqrt{1999}}\)
Tính \(A=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{1999\sqrt{1998}+1998\sqrt{1999}}\)
Tính
N=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{1999\sqrt{1998}+1998\sqrt{1999}}\)
giúp mình với
Cho
S = \(\frac{1}{\sqrt{1.2014}}+\frac{1}{\sqrt{2.2013}}+......+\frac{1}{\sqrt{k.\left(2014-k+1\right)}}+.....+\frac{1}{\sqrt{2014.1}}\)
Hãy so sánh S với \(2.\frac{2014}{2015}\)
Chứng minh:\(\frac{1}{2\sqrt{1}}+\frac{1}{3\sqrt{2}}+\frac{1}{4\sqrt{3}}+...+\frac{1}{1999\sqrt{1998}}< 2\)<2
chứng minh
\(1998< 1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{1000000}}< 1999\)
