\(\frac{n^3-1}{n^3+1}=\frac{\left(n-1\right)\left(n^2+n+1\right)}{\left(n+1\right)\left(n^2-n+1\right)}=\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)

\(\Rightarrow u_n=\frac{1.\left(3^2-3+1\right)}{3.\left(2^2-2+1\right)}.\frac{2\left(4^2-4+1\right)}{4.\left(3^2-3+1\right)}.\frac{3\left(5^2-5+1\right)}{5\left(4^2-4+1\right)}...\frac{\left(n-1\right)\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(n+1\right)\left(n^2-n+1\right)}\)

\(\Rightarrow u_n=\frac{1.2.\left[\left(n+1\right)^2-\left(n+1\right)+1\right]}{\left(2^2-2+1\right).n\left(n+1\right)}=\frac{2n^2+2n+2}{3n^2+3n}\)

\(\Rightarrow lim\left(u_n\right)=lim\frac{2n^2+2n+2}{3n^2+3n}=\frac{2}{3}\)

a/

\(u_n=\dfrac{1}{\left(2-1\right)\left(2+1\right)}+\dfrac{1}{\left(3-1\right)\left(3+1\right)}+...+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(u_n=\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(n-2\right)n}+\dfrac{1}{\left(n-1\right)\left(n+1\right)}\)

\(u_n=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{n-2}-\dfrac{1}{n}+\dfrac{1}{n-1}-\dfrac{1}{n+1}\right)\)

\(u_n=\dfrac{1}{2}\left(1+\dfrac{1}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)=\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\)

\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{2}\left(\dfrac{3}{2}-\dfrac{1}{n}-\dfrac{1}{n+1}\right)\right)=\dfrac{1}{2}.\dfrac{3}{2}=\dfrac{3}{4}\)

b/ \(u_n=\dfrac{1}{1^2+3}+\dfrac{1}{2^2+6}+...+\dfrac{1}{n^2+3n}=\dfrac{1}{1.4}+\dfrac{1}{2.5}+...+\dfrac{1}{n\left(n+3\right)}\)

\(u_n=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{n}-\dfrac{1}{n+3}\right)\)

\(u_n=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\)

\(\Rightarrow lim\left(u_n\right)=lim\left(\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{n+1}-\dfrac{1}{n+2}-\dfrac{1}{n+3}\right)\right)\)

\(\Rightarrow lim\left(u_n\right)=\dfrac{1}{3}\left(1+\dfrac{1}{2}+\dfrac{1}{3}\right)=\dfrac{11}{18}\)

1)

Vì -1\(\le\) sin(5n)\(\le\) 1

Nên \(\lim\limits_{n\rarr+\infty}\left(\frac{\sin\left(5n\right)}{3n}-2\right)\) = -2

2)

\(-1\le\cos2n\le1\)

Có \(\lim\limits_{n\rarr+\infty}\left(5-\frac{\left(n^2\cos2n\right)}{n^2+1}\right)\)

= \(\lim\limits_{n\rarr+\infty}5-\frac{\left(\cos2n\right)}{1+\frac{1}{n^2}}\) =A => A nhận các giá trị trong đoạn [4;6]

3)

Có \({\sum_1^{+\infty}\frac{\frac{n}{2}}{n^2+1}}\) =\(\) \(\frac{\frac12+\frac12\left(n-1\right)}{n^2+1}\) nên lim của nó =0

4)

4)

\(\sum_1^{+\infty}\) \(\frac{\left(-1\right)^{n+1}}{2^{n}}\) =\(\lim\limits_{n\rarr+\infty}\) \(\frac{\frac12\left(1-\left(-\frac12\right)^{n}\right)}{1-\frac{-1}{2}}\) =\(\frac13\)

5)

\(\lim\limits_{n\rarr+\infty}\) \(\frac{n-2\sqrt{n}\sin2n}{2n}\) =\(\frac12\)