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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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    $\begingroup$ Numerically, the answer looks like $\frac19$. $\endgroup$ – Simply Beautiful Art Feb 6 '17 at 15:13
  • $\begingroup$ dominated convergence theorem will finish this one off quickly $\endgroup$ – tired Feb 6 '17 at 15:25
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    $\begingroup$ Note that $4I_n+5I_{n-1}=\frac{1}{n}$ and hence if $nI_n$ tends to a limit it must be $1/9$. $\endgroup$ – Paramanand Singh Feb 6 '17 at 15:42
  • $\begingroup$ @ParamanandSingh +1 Wow! Nice comment, thank you for sharing. $\endgroup$ – MrYouMath Feb 6 '17 at 16:04
  • $\begingroup$ @MrYouMath: I still need to show that $nI_n$ converges. $\endgroup$ – Paramanand Singh Feb 6 '17 at 16:06
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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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    $\begingroup$ Simple is better. +1 $\endgroup$ – Paramanand Singh Feb 6 '17 at 18:24
  • $\begingroup$ @ParamanandSingh As my username promotes. $\endgroup$ – Simply Beautiful Art Feb 7 '17 at 0:52
  • $\begingroup$ Why was this downvoted? :( $\endgroup$ – S.C.B. Feb 7 '17 at 2:04
  • $\begingroup$ @S.C.B.: it happens to me quite often. I guess it is a retaliation for the times I voted for closing some question. I cannot do anything about it, and if it is not a serial thing, I cannot even complain with the mods. I guess I just have to get used to it. $\endgroup$ – Jack D'Aurizio Feb 7 '17 at 20:36
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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:

$$I_n=\int_0^1{1\over 5}\sum_{i=0}^\infty (-1)^i\left({4\over 5}\right)^ix^{n+i}$$

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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$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\mrm{g}\pars{n,x} \equiv {\pars{1 - x}^{n} \over 9 - 4x}$

$\ds{\mrm{g}\pars{n,x} \equiv {\pars{1 - x}^{n} \over 9 - 4x}}$


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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  • $\begingroup$ Felix, how did you arrive at the penultimate step? $\endgroup$ – Mark Viola Feb 7 '17 at 18:24
  • $\begingroup$ @Dr.MV $$ n\int_{0}^{1}{\exp\left(n\ln\left(1 - x\right)\right)\over 1 - 4x/9}\,\mathrm{d}x \sim n\int_{0}^{\infty}\mathrm{e}^{-nx}\left(1 + {4 \over 9}\,x\right)\,\mathrm{d}x = 1 + {4 \over 9}\,{1 \over n}\quad\mbox{as }\ n \to \infty $$ This is the Laplace Method. $\endgroup$ – Felix Marin Feb 7 '17 at 20:24
  • $\begingroup$ It might help other readers to add this step.. (+1) $\endgroup$ – Mark Viola Feb 7 '17 at 20:41
  • $\begingroup$ @Dr.MV Thanks. I'll do it. $\endgroup$ – Felix Marin Feb 7 '17 at 21:29

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