Ta có: \(\frac{1}{2}>\frac{1}{100}\)
\(\frac{1}{3}>\frac{1}{100}\)
\(\frac{1}{4}>\frac{1}{100}\)
.....
\(\frac{1}{99}>\frac{1}{100}\)
nên \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{99}+\frac{1}{100}>99.\frac{1}{100}=\frac{99}{100}\)
Chứng minh rằng :
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+\frac{4}{5}+...+\frac{99}{100}\)
Chứng minh rằng
\(100\)\(-(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100})=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)\(\frac{99}{100}\)
Chứng minh
100-(1-\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}\))= \(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
chứng minh rằng:
\(100-\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{100}\right)=\frac{1}{2}+\frac{1}{3}+\frac{3}{4}+....+\frac{99}{100}\)
Chứng minh rằng:
a,\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
b,\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
giúp minh với
Chứng minh rằng:
\(100-\left(1+\frac{1}{3}+...+\frac{1}{100}\right)=\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{99}{100}\)
Bài 5 chứng minh: \(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{3}{16}\)
Chứng minh rằng :
\(\left(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{99}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right)=\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+\frac{1}{54}+...+\frac{1}{100}\)
\(\frac{1}{2}-\frac{-2}{2^2}+\frac{3}{2^3}-\frac{4}{2^4}+\frac{4}{2^5}+...+\frac{99}{2^{99}}-\frac{100}{2^{100}}< \frac{2}{9}\)
Chứng minh
