Let $(I_n)_{n \geq 1}$ be a sequence such that: $$I_n = \int_0^1 \frac{x^n}{4x + 5} dx$$ Evaluate the following limit: $$\lim_{n \to \infty} nI_n$$
All I've been able to find is that $(I_n)$ is decreasing and converges to $0$.
Thank you!
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It is enough to apply integration by parts: $$ n I_n = \int_{0}^{1}(n x^{n-1})\frac{x\,dx}{4x+5} = \color{red}{\frac{1}{9}}-\int_{0}^{1}\frac{5x^n}{(4x+5)^2}\,dx$$ and notice that: $$ \left|\int_{0}^{1}\frac{5x^n}{(4x+5)^2}\,dx\right|\leq \frac{1}{5}\int_{0}^{1}x^n\,dx = \frac{1}{5n+5}.$$ |
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Hint: Expand the integrand by using the geometric series: $$\frac{x^n}{5(1+4x/5)}.$$ It is uniformly convergent on $|x|\leq 5/4$ and integrate the resulting polynomial. |
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HINT: If $\displaystyle I_m=\int_0^1\dfrac{x^m}{4x+5}dx,$ $\displaystyle4I_{n+1}=\int_0^1\dfrac{4x^{n+1}}{4x+5}dx=\int_0^1\dfrac{x^n(4x+5)-5x^n}{4x+5}dx=\int_0^1x^n\ dx-5I_n$ Now $\displaystyle4\cdot\dfrac{I_{n+1}}{n+1}+\dfrac{5n}{n+1}\cdot\dfrac{I_n}n=\dfrac{\int_0^1x^n\ dx}{n+1}$ Finally, $$\lim_{n\to\infty}\dfrac{I_{n+1}}{n+1}=\lim_{n\to\infty}\dfrac{I_n}n$$ |
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Expand the power series:
Since the function converges uniformly on $[0,1]$ and absolutely, we can pull the integral inside and evaluate and we get $$I_n = {1\over 5}\sum_{i=0}^\infty(-1)^i\left({4\over 5}\right)^i{1\over n+i+1}$$ Multiply by $n$ and we get $$nI_n = {1\over 5}\sum_{i=0}^n(-1)^{i}\left({4\over 5}\right)^{i}{n\over n+i+1}={1\over 5}\sum_{i=0}^\infty (-1)^{i}\left({4\over 5}\right)^{i}\left(1-{i+1\over n+i+1}\right)$$ We can evaluate the first sum easily enough $${1\over 5}\sum_{i=0}^\infty (-1)^{i}\left({4\over 5}\right)^i={1\over 5+4}={1\over 9}$$ Now for the second term we can use absolute convergence to take the limit inside the sum and we get: $$\lim_{n\to\infty} \sum_{i=0}^\infty(-1)^i\left({4\over 5}\right)^i{i+1\over n+i+1}=0=\sum_{i=0}^\infty(-1)^i\left({4\over 5}\right)^i(i+1)\cdot\lim_{n\to\infty}{1\over n+i+1}=0$$ so the end result is just ${1\over 9}$. |
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This is an application of Laplace Method: \begin{align} \lim_{n \to \infty}\pars{n\int_{0}^{1}{x^{n} \over 4x + 5}\,\dd x} & = {1 \over 9}\lim_{n \to \infty}\bracks{n \int_{0}^{1}{\pars{1 - x}^{n} \over 1 - 4x/9}\,\dd x} = {1 \over 9}\lim_{n \to \infty}\pars{n\int_{0}^{\infty}\expo{-nx}\,\dd x} \\[5mm] & = \bbx{\ds{1 \over 9}} \end{align} |
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