Tính $\lim_{n\rightarrow +\infty}\frac{I_{n+1}}{I_n}$

Cho $I_n=\int_{0}^{1}x^2(1-x^2)^ndx, n\in N*$. Tính $\lim_{n\rightarrow +\infty}\frac{I_{n+1}}{I_n}$
Lời giải
Ta có: $${I_{n + 1}} = \int\limits_0^1 {{x^2}{{\left( {1 - {x^2}} \right)}^{n+1}}dx = \int\limits_0^1 {{x^2}\left( {1 - {x^2}} \right){{\left( {1 - {x^2}} \right)}^n}} } dx = \int\limits_0^1 {{x^2}{{\left( {1 - {x^2}} \right)}^n}dx - } \int\limits_0^1 {{x^4}{{\left( {1 - {x^2}} \right)}^n}dx}$$
$$= {I_n} + \frac{1}{2}\int\limits_0^1 {{x^3}} {\left( {1 - {x^2}} \right)^n}d\left( {1 - {x^2}} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
Xét tích phân: $${J_n} = \int\limits_0^1 {{x^3}} {\left( {1 - {x^2}} \right)^n}d\left( {1 - {x^2}} \right)$$
Đặt $$\left\{ \begin{array}{l} u = {x^3}\\ dv = {\left( {1 - {x^2}} \right)^n}d\left( {1 - {x^2}} \right) \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = 3{x^2}dx\\ v = \frac{{{{\left( {1 - {x^2}} \right)}^{n + 1}}}}{{n + 1}} \end{array} \right.$$
Khi đó: $${J_n} = \left. {{x^3}\frac{{{{\left( {1 - {x^2}} \right)}^{n + 1}}}}{{n + 1}}} \right|_0^1 - \frac{3}{{n + 1}}\int\limits_0^1 {{x^2}{{\left( {1 - {x^2}} \right)}^{n + 1}}dx =  - } \frac{3}{{n + 1}}{I_{n + 1}}\,\,\,\,\,\,\,\,\,\,\,\,(2)$$
Thay $(2)$ vào $(1)$, ta được: $${I_{n + 1}} = {I_n} - \frac{1}{2}.\frac{3}{{n + 1}}{I_{n + 1}} \Rightarrow \frac{{{I_{n + 1}}}}{{{I_n}}} = \frac{{2n + 2}}{{2n + 5}} \Rightarrow \mathop {\lim }\limits_{n \to \infty } \frac{{{I_{n + 1}}}}{{{I_n}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{2n + 2}}{{2n + 5}} = 1$$