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If $\displaystyle I_{n}=\int^{1}_{0}x^2(1-x^2)^ndx,$ Then $\displaystyle \lim_{n\rightarrow \infty}\frac{I_{n+1}}{I_{n}}$ is

plan

$$I_{n}=-x\cdot \frac{(1-x^2)^{n+1}}{2(n+1)}\bigg|^{1}_{0}+\frac{1}{2(n+1)}\int^{1}_{0}(1-x^{2})^{n+1}dx$$

$$I_{n}=\frac{1}{2(n+1)}\int^{1}(1-x^2)^{n+1}dx$$

help me please to solve it

jacky
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3 Answers3

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hint...you need to get a recurrence relation for $I_n$, so write $$(1-x^2)^{n+1}=(1-x^2)(1-x^2)^n$$

David Quinn
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$I_n=\frac {x^{3}} 3 (1-x^{2})^{n}|_0^{1}+\frac {2n} 3\int_0^{1} x^{4}(1-x^{2})^{n-1}dx=\frac {2n} 3\int_0^{1} x^{4}(1-x^{2})^{n-1}dx$. Now write $x^{4}$ as $x^{2}-x^{2}(1-x^{2})$ to derive the formula $I_n=\frac {2n} 3 I_{n-1}-\frac {2n} 3I_n$. This gives $I_n=\frac {2n} {2n+3} I_{n-1}$ from which it follows that the required limit is $1$.

Kavi Rama Murthy
  • 311,013
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Using your intermediate result for $I_n$ you may proceed as follows:

Setting $\boxed{J_n = \int_0^1(1-x^2)^n dx}$, it is quickly verified that $$J_{n+1} = J_{n} - I_n \Leftrightarrow I_n = J_{n} - J_{n+1}\quad (\star )$$

You calculated already: $$I_n = \frac{1}{2(n+1)}J_{n+1}$$

Plugging your result into $(\star)$ you get $$\frac{1}{2(n+1)}J_{n+1} = J_{n} - J_{n+1} \Leftrightarrow \frac{J_{n+1}}{J_{n}} = \frac{2(n+1)}{2(n+1)+1} \quad (\star \star)$$

It follows $$\frac{I_{n+1}}{I_n}= \frac{J_{n+1} - J_{n+2}}{J_{n} - J_{n+1}}= \frac{\frac{1}{2(n+2)}J_{n+2}}{\frac{1}{2(n+1)}J_{n+1}}=\frac{n+1}{n+2}\cdot \frac{J_{n+2}}{J_{n+1}}$$ $$\stackrel{(\star \star)}{=} \frac{n+1}{n+2}\cdot \frac{2(n+2)}{2(n+2)+1}\stackrel{n \to \infty}{\longrightarrow}1$$

trancelocation
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