CMR, ∀n ≥ 1, n ∈ N : \(\dfrac{1}{2}\)+\(\dfrac{1}{3\sqrt{2}}\)+\(\dfrac{1}{4\sqrt{3}}\)+....+ \(\dfrac{1}{\left(n+1\right)\sqrt{n}}\)<2
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10 tháng 9 2017
\(\dfrac{1}{\sqrt{n}.\sqrt{n+1}.\left(\sqrt{n+1}+\sqrt{n}\right)}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{1}{\sqrt{n+1}+\sqrt{n}}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\dfrac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\dfrac{1}{\sqrt{n}.\sqrt{n+1}}.\left(\sqrt{n+1}-\sqrt{n}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n}.\sqrt{n+1}}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 6 2021
\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}\)
\(=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
Do đó:
\(VT=\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)
\(VT=1-\dfrac{1}{\sqrt{n+1}}< 1\) (đpcm)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(a=\lim\sqrt{n^3}\sqrt{\dfrac{1}{n^3}+\dfrac{2}{n^2}-1}=\infty.\left(-1\right)=-\infty\)
\(b=\lim\left(\sqrt{n^2+2n+3}-n+n-\sqrt[3]{n^2+n^3}\right)\)
\(=\lim\dfrac{2n+3}{\sqrt{n^2+2n+3}+n}+\lim\dfrac{-n^2}{n^2+n\sqrt[3]{n^2+n^3}+\sqrt[3]{\left(n^2+n^3\right)^2}}\)
\(=\lim\dfrac{2+\dfrac{3}{n}}{\sqrt{1+\dfrac{2}{n}+\dfrac{3}{n^2}}+1}+\lim\dfrac{-1}{1+\sqrt[3]{\dfrac{1}{n}+1}+\sqrt[3]{\left(\dfrac{1}{n}+1\right)^2}}=\dfrac{2}{2}-\dfrac{1}{3}=\dfrac{2}{3}\)
\(c=\lim\dfrac{\left(\dfrac{2}{\sqrt{n}}+\dfrac{1}{n}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{3}{n}\right)}{\left(1+\dfrac{1}{n}\right)\left(1+\dfrac{2}{n}\right)}=\dfrac{0.0}{1.1}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
20 tháng 1 2021
\(d=\lim\dfrac{4-3\left(\dfrac{2}{4}\right)^n}{9.\left(\dfrac{3}{4}\right)^n+\left(\dfrac{2}{4}\right)^n}=\dfrac{4}{0}=+\infty\)
\(e=\lim\dfrac{7-25\left(\dfrac{5}{7}\right)^n+3.\left(\dfrac{1}{7}\right)^n}{12.\left(\dfrac{6}{7}\right)^n-\left(\dfrac{3}{7}\right)^n+3\left(\dfrac{1}{7}\right)^n}=\dfrac{7}{0}=+\infty\)
\(f=\lim\dfrac{n^4-4n^6}{n\left(\sqrt{n^4+1}+\sqrt{4n^6+1}\right)}=\lim\dfrac{\dfrac{1}{n^2}-6}{\sqrt{\dfrac{1}{n^6}+\dfrac{1}{n^{10}}}+\sqrt{\dfrac{4}{n^4}+\dfrac{1}{n^{10}}}}=\dfrac{-6}{0}=-\infty\)
17 tháng 9 2018
Ta có : \(\dfrac{1}{\left(n+1\right)\sqrt{n}}=\dfrac{\sqrt{n}}{n\left(n+1\right)}=\sqrt{n}\left(\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\sqrt{n}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\left(\dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}\right)=\left(1+\dfrac{\sqrt{n}}{\sqrt{n+1}}\right)\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Áp dụng điều này vào bài toán , ta có :
\(\left\{{}\begin{matrix}\dfrac{1}{2\sqrt{1}}< 2\left(1-\dfrac{1}{\sqrt{2}}\right)=2-\sqrt{2}\\\dfrac{1}{3\sqrt{2}}< 2\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\right)=\sqrt{2}-\dfrac{2}{\sqrt{3}}\\....\\\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{2}{\sqrt{n}}-\dfrac{2}{\sqrt{n+1}}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{2\sqrt{1}}+\dfrac{1}{2\sqrt{3}}+...+\dfrac{1}{\left(n+1\right)\sqrt{n}}< 2-\dfrac{2}{\sqrt{n+1}}< 2\) ( Sửa đề ^-^ )