=1/1.2+1/1.2.3+.............1/1.2.3.4. ......... .2019
=1-1/2+1/2-1/3+1/3-.............-2019
=1-1/2019
=2018/2019
Vay 1/1.2+1/1.2.3+.............1/1.2.3.4. ......... .2019>1/2
Cho tổng \(T=\dfrac{2}{2^1}+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{2020}{2^{2019}}+\dfrac{2021}{2^{2020}}\)
So sánh T với 3
So sánh : \(A=\dfrac{2019^{2020}+1}{2019^{2019}-1}\) và \(B=\dfrac{2019^{2019}+1}{2019^{2018}-1}\)
\(S=\dfrac{2}{2021+1}+\dfrac{2^2}{2021^2+1}+\dfrac{2^3}{2021^{2^2}+1}+...+\dfrac{2^{n+1}}{2021^{2^n}+1}+...+\dfrac{2^{2021}}{2021^{2^{2020}}+1}\)
So sánh S với \(\dfrac{1}{1010}\)
So sánh A và B :
\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{100}}\)
\(B=\dfrac{1}{2}\)
Cho \(A=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2019}\)
Chứng minh A ko phải là số tự nhiên
Cho A = \(\dfrac{1}{5^2}+\dfrac{1}{6^2}+\dfrac{1}{7^2}+.....+\dfrac{1}{2019^2}\)
Chứng minh rằng \(\dfrac{20}{101}< A< \dfrac{1}{4}\)
ko tinh
a,\(2020-\dfrac{1}{3^2}-\dfrac{1}{4^2}-...-\dfrac{1}{2019^2}\)
b,\(\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{17}\)
so sánh \(\dfrac{1}{2^{500}}\) và \(\dfrac{1}{5^{200}}\)
So sánh A =\(\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\)với 2 ta được A...2
