\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+..+\frac{1}{3^{99}}+\frac{1}{3^{100}}\) và \(\frac{1}{2}\)
So sánh
Cho \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)
so sánh với 2
so sánh :
A = \(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.......+\frac{1}{3^{99}}\) với B = \(\frac{1}{2}\)
Hãy so sánh : \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+....+\frac{99}{100!}\) và 1
So sánh:
\(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}}với\frac{1}{2}\)
Chứng minh rằng:
a. \(\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+\frac{4}{3^5}+...+\frac{99}{3^{100}}+\frac{100}{3^{101}}< \frac{1}{4}\)
b.\(\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\frac{1}{32}-\frac{1}{64}< \frac{1}{3}\)
c.\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}< \frac{1}{16}\)
d. \(\frac{1}{5^2}-\frac{2}{5^3}+\frac{3}{5^4}-\frac{4}{5^5}+...+\frac{99}{5^{100}}-\frac{100}{5^{101}}< \frac{1}{36}\)
\(A=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3A=1-\frac{2}{3}+\frac{3}{3^2}-...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow3A+A=\left(...\right)+\left(...\right)\)
\(\Rightarrow4A=1-\frac{1}{3}+\frac{1}{3^2}-...-\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow3.4A=3-1+\frac{1}{3}-...-\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow12A+4A=\left(...\right)+\left(...\right)\)
\(\Rightarrow16A=3-\frac{101}{3^{99}}-\frac{100}{3^{100}}< 3\)
\(\Rightarrow A< \frac{3}{16}\)
so sánh \(\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.............+\frac{1}{3^{99}}\right)và\frac{1}{2}\)
Cho \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{100}};B=\frac{1}{2}\).so sánh A và B