\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{n\left(n+1\right)}\)

= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\)

= 1 - \(\dfrac{1}{n+1}\) = \(\dfrac{n}{n+1}\)

\(A=\dfrac{3}{\left(1\cdot2\right)^2}+\dfrac{5}{\left(2\cdot3\right)^2}+\dfrac{7}{\left(3\cdot4\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(A=\dfrac{3}{1\cdot4}+\dfrac{5}{4\cdot9}+\dfrac{7}{9\cdot16}+...+\dfrac{2n+1}{n^2\cdot\left(n^2+2n+1\right)}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

\(A=1-\dfrac{1}{n^2+2n+1}\)

\(A=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Đặt \(S=\dfrac{1}{\sqrt{n^3+1}}+\dfrac{1}{\sqrt{n^3+2}}+...+\dfrac{1}{\sqrt{n^3+n}}\)

\(n^3+n>...>n^3+2>n^3+1\)

\(\Rightarrow\dfrac{n}{\sqrt{n^3+n}}< S< \dfrac{n}{\sqrt{n^3+1}}\)

Mà \(\lim\left(\dfrac{n}{\sqrt{n^3+1}}\right)=\lim\left(\dfrac{n}{\sqrt{n^3+n}}\right)=0\)

\(\Rightarrow\lim\left(S\right)=0\)

\(A=\dfrac{3}{\left(1.2\right)^2}+\dfrac{5}{\left(2.3\right)^2}+...+\dfrac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(=\dfrac{3}{1.4}+\dfrac{5}{4.9}+...+\dfrac{2n+1}{n^2\left(n^2+2n+1\right)}\)

\(=\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+...+\dfrac{1}{n^2}-\dfrac{1}{n^2+2n+1}\)

\(=1-\dfrac{1}{n^2+2n+1}\)

\(=\dfrac{n^2+2n}{n^2+2n+1}=\dfrac{n\left(n+2\right)}{\left(n+1\right)^2}\)

Xét thừa số tổng quát:

\(\dfrac{k}{\left(\dfrac{k-1}{2}.\dfrac{k+1}{2}\right)^2}\)\(=\dfrac{k}{\left(\dfrac{\left(k-1\right)\left(k+1\right)}{4}\right)^2}=\dfrac{k}{\left(\dfrac{\left(k-1\right)\left(k+1\right)}{4}\right)^2}\)

\(=\dfrac{k}{\dfrac{\left[\left(k-1\right)\left(k+1\right)\right]^2}{16}}=\dfrac{k}{\dfrac{\left(k^2-1\right)^2}{16}}=\dfrac{16k}{\left(k^2-1\right)^2}\)

Thay \(k=3;5;....2n+1\) ta được:

\(\dfrac{16.3}{\left(3^2-1\right)^2}+\dfrac{16.5}{\left(5^2-1\right)^2}+....+\dfrac{16.n}{\left(n^2-1\right)^2}\)

\(=16.\left(\dfrac{3}{\left(3^2-1\right)^2}+\dfrac{5}{\left(5^2-1\right)^2}+...+\dfrac{n}{\left(n^2-1\right)^2}\right)\)

\(=16.\left(\dfrac{3}{\left[\left(3-1\right)\left(3+1\right)\right]^2}+\dfrac{5}{\left[\left(5-1\right)\left(5+1\right)\right]^2}+...+\dfrac{n}{\left[\left(n-1\right)\left(n+1\right)\right]^2}\right)\)

\(=16.\left(\dfrac{3}{4.16}+\dfrac{5}{16.36}+...+\dfrac{n}{\left(n-1\right)^2.\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{12}{4.16}+\dfrac{20}{16.36}+...+\dfrac{4n}{\left(n-1\right)^2.\left(n+1\right)^2}\right)\)

\(=4.\left(\dfrac{1}{4}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{36}+...+\dfrac{1}{\left(n-1\right)^2}-\dfrac{1}{\left(n+1\right)^2}\right)\)

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

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

\(=4.\left(\dfrac{\left(n+1\right)^2-4}{4\left(n+1\right)^2}\right)=\dfrac{4\left(n+1\right)^2-16}{4\left(n+1\right)^2}\)

\(=\dfrac{4\left[\left(n+1\right)^2-4\right]}{4\left(n+1\right)^2}=\dfrac{\left(n+1\right)^2-4}{\left(n+1\right)^2}\)

Chúc bạn học tốt!!!

\(a=lim\dfrac{\left(\dfrac{2}{6}\right)^n+1-\dfrac{1}{4}\left(\dfrac{4}{6}\right)^n}{\left(\dfrac{3}{6}\right)^n+6}=\dfrac{1}{6}\)

\(b=\lim\dfrac{\left(n+1\right)^2}{3n^2+4}=\lim\dfrac{n^2+2n+1}{3n^2+4}=\lim\dfrac{1+\dfrac{2}{n}+\dfrac{1}{n^2}}{3+\dfrac{4}{n^2}}=\dfrac{1}{3}\)

\(c=\lim\dfrac{n\left(n+1\right)}{2\left(n^2-3\right)}=\lim\dfrac{n^2+n}{2n^2-6}=\lim\dfrac{1+\dfrac{1}{n}}{2-\dfrac{6}{n^2}}=\dfrac{1}{2}\)

\(d=\lim\left[1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n}-\dfrac{1}{n+1}\right]=\lim\left[1-\dfrac{1}{n+1}\right]=1\)

\(e=\lim\dfrac{1}{2}\left[1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{2n-1}-\dfrac{1}{2n+1}\right]\)

\(=\lim\dfrac{1}{2}\left[1-\dfrac{1}{2n+1}\right]=\dfrac{1}{2}\)

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

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

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

Lời giải:

Xét số hạng tổng quát:

\(\frac{2n+1}{[n(n+1)]^2}=\frac{1}{n(n+1)}.\frac{2n+1}{n(n+1)}=\frac{n+1-n}{n(n+1)}.\frac{n+(n+1)}{n(n+1)}\)

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

Do đó:

\(S=\frac{3}{(1.2)^2}+\frac{5}{(2.3)^2}+....+\frac{2n+1}{[n(n+1)]^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{(n+1)^2}\)

\(=1-\frac{1}{(n+1)^2}\)

a) = 1-1/2+1/2-1/3+...+1/99-1/100 =1 - 1/100 = 99/100