Ta có \(1-\frac{1}{k^2}=\frac{\left(k-1\right)\left(k+1\right)}{k^2}\)
=> \(limS=lim\frac{1.3}{2^2}.\frac{2.4}{3^2}...\frac{\left(n-1\right)\left(n+1\right)}{n^2}=lim\frac{n+1}{2n}=\frac{1}{2}\)
\(\lim\limits\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+...+n}\right)\)
a; lim\(\frac{\sqrt{6n^4+n+1}}{2n^2+1}\)
b; lim \(\frac{\left(n+1\right)\left(2n+1\right)^2\left(3n+1\right)^3}{n^2\left(n+2\right)^2\left(1-3n\right)^2}\)
\(\lim\limits\left(\frac{1}{2.4}+\frac{1}{5.7}+\frac{1}{8.10}+...+\frac{1}{\left(3n-1\right)\left(3n+1\right)}\right)\)
lim \(\frac{\left(2n^2-3n+5\right)\left(2n+1\right)}{\left(4-3n\right)\left(2n^2+n+1\right)}\)
lim \(\frac{\sqrt{n^4+1}}{n}-\frac{\sqrt{4n^6+2}}{n^2}\)
lim \(\frac{2n+3}{\sqrt{9n^2+3}-\sqrt[3]{2n^2-8n^3}}\)
Tính \(\lim\limits\left[\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{\left(2n-1\right)\left(2n+1\right)}\right]\)
Cho \(0< \left|a\right|,\left|b\right|< 1\). Khi đó \(\lim\limits\frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}\)=
tìm giới hanjn
1) lim \(\frac{\left(-1\right)^n}{n-3}\)
2) lim \(\frac{n\left(sin\left(pi.n^2\right)\right)}{n^2+3n-2}\)
\(\lim\limits\left[\left(1-n\right)\left(\sqrt{n^2-6n}-\sqrt[3]{n^3-27n^2}\right)\right]\)
lim (\(\frac{1}{2}\)+ \(\frac{\left(-1\right)^n.\left(\sqrt{3}\right)^n}{3.2^{n+1}}\))