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Given $I_n=\displaystyle\int_0 ^1{\frac{x^{n+1}}{x+3}}\,dx$ for $n\in\mathbb N$, prove that $$\lim_{n \to \infty} nI_n=\frac{1}{4}$$

Here is what the limit looks like:

$$\lim_{n \to \infty} n\int_0 ^1{\frac{x^{n+1}}{x+3}}\,dx$$

I am not sure how to go about this exercise, but I tried solving the integral with no success. I integrated by parts once choosing $u = x^{n+1}$ and $v = \frac{1}{x+3}$ and this is what I got. $$I_n=\ln 4-(n+1)\int_0^1{x^n\ln(x+3)}\,dx$$
I don't know how to proceed. I was asked previously to prove that $I_{n+1}+3I_n=\frac{1}{n+2}$, do you think this can be used somehow? I tried solving for $I_n$ but I don't see how it can help because of the $I_{n+1}$ that is left over? Help me out, please!

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Notice that

$$I_{n+1}-I_n = \int_0^1 \frac{x^{n+1}(x-1)}{x+3} \text{d}x \leq 0$$

so the sequence $\left(I_n\right)_{n\ge 1}$ is decreasing. Also:

$$3I_n+I_{n+1} = \int_0^1 \frac{3x^{n+1}+x^{n+2}}{x+3}\text{d}x = \int_0^1 x^{n+1} \text{d}x = \frac{1}{n+2}$$

Now since the sequence is decreasing we have $I_{n+1} \leq I_{n}$. Therefore:

$$3I_n+I_{n+1}\leq 3I_n+I_n= 4I_n\ \ \ \ \ \ \ \ \ \ (1)$$

Similarly, because $I_n \leq I_{n-1} \Rightarrow 3I_n \leq 3I_{n-1}$, we have:

$$4I_n = 3I_n + I_n \leq 3I_{n-1}+I_n\ \ \ \ \ \ \ \ \ \ (2)$$

Chaining $(1)$ and $(2)$, we get:

$$3I_{n}+I_{n+1} \leq 4I_n \leq 3I_{n-1}+I_{n}$$

or

$$\frac{1}{n+2} \leq 4I_n \leq \frac{1}{n+1}$$

which implies

$$\frac{n}{4(n+2)} \leq nI_n \leq \frac{n}{4(n+1)}$$

From the squeeze theorem, it follows that:

$$\lim_{n\to \infty} nI_n = \frac{1}{4}$$

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  • $\begingroup$ This was really useful, thanks a lot! $\endgroup$ – Radu Gabriel Jan 26 '20 at 11:35
  • $\begingroup$ I have a couple of questions please. You said "Therefore (from the decreasing monotony): $3I_{n+1}+I_n \leq 4I_n \leq 3I_{n-1}+I_{n}$. I see how you got to this expression, but the next one I just can't figure out. $\frac{1}{n+2} \leq 4I_n \leq \frac{1}{n+1}$. I know that previous to that we had $3I_n+I_{n+1} = \frac{1}{n+2}$ but how is $3I_n+I_{n+1}$ the same as $3I_{n+1}+I_n$? $\endgroup$ – Radu Gabriel Jan 26 '20 at 12:11
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    $\begingroup$ Sorry, it's a typo on my part, it should be $3I_{n}+I_{n+1} $ on the left side. I edited. $\endgroup$ – LHF Jan 26 '20 at 12:16
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    $\begingroup$ The left inequality is equivalent (after subtracting $3I_n$) with $I_{n+1} \leq I_n$, and the right inequality is equivalent (after subtracting $I_n$) with $3I_n \leq 3I_{n-1}$ or $I_n \leq I_{n-1}$. Both of these inequalities are true because $I_n$ is decreasing, id est $I_{n+1}\leq I_{n} \leq I_{n-1}$. I'm here if there's something else that's not clear. $\endgroup$ – LHF Jan 26 '20 at 12:30
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    $\begingroup$ That's because I have already shown the recurrence formula $3I_n+I_{n+1}=\dfrac{1}{n+2}$ is true for any integer $n$. So just set $n \to n-1$ in the recurrence formula to get $3I_{n-1}+I_n = \dfrac{1}{n+1}$. Or you can repeat the process in the second line: $3I_{n-1}+I_{n} = \int_0^1 \frac{3x^{n}+x^{n+1}}{x+3}\text{d}x = \int_0^1 x^{n} \text{d}x = \frac{1}{n+1}$ $\endgroup$ – LHF Jan 26 '20 at 13:06
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Put $t=x^{n}$. We get $nI_n=\int_0^{1} \frac {y^{1/n}} {{y^{1/n}+3}} \to \frac 1 4$ because $\frac {y^{1/n}} {{y^{1/n}+3}} \to \frac 1 4$ and $0 <\frac {y^{1/n}} {{y^{1/n}+3}} <1$. [Use DCT to justify interchange of limit and integral].

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You can also calculate the limit directly without any recursions using partial integration as follows:

\begin{eqnarray} n\int_0 ^1{\frac{x^{n+1}}{x+3}}dx & = & \int_0^1 \frac{(n+1)x^n\cdot x - x^{n+1}}{x+3}dx\\ & = & \underbrace{\int_0^1 (n+1)x^n \left(1-\frac 3{x+3}\right)dx}_{J_n} - \underbrace{\int_0^1 \frac{x^{n+1}}{x+3}dx}_{\leq \frac 1{3(n+2)}\stackrel{n\to\infty}{\rightarrow}0}\\ \end{eqnarray}

So, it is enough to evaluate $J_n$ using partial integration: \begin{eqnarray} J_n & = & \underbrace{\left[x^{n+1}\left(1-\frac 3{x+3}\right)\right]_0^1}_{=\frac 14} - \underbrace{\int_0^1x^{n+1}\frac{3}{(x+3)^2}dx}_{\leq \frac 1{3(n+2)}\stackrel{n\to\infty}{\rightarrow}0} \\ & \stackrel{n\to\infty}{\rightarrow} & \frac 14 \end{eqnarray}

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$$ \begin{align} \lim_{n\to\infty}n\int_0^1\frac{x^{n+1}}{x+3}\,\mathrm{d}x &=\lim_{n\to\infty}\underbrace{\vphantom{\frac1{x^{\frac12}}}\ \ \frac{n}{n+2}\ \ }_1\underbrace{\int_0^1\frac1{x^{\frac1{n+2}}+3}\,\mathrm{d}x}_{1/4}\tag1\\ &=\frac14\tag2 \end{align} $$ Explanation:
$(1)$: substitute $x\mapsto x^{\frac1{n+2}}$
$(2)$: Dominated Convergence

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By writing $n I_n = \frac{n}{n+2} \cdot (n+2)I_n$ and noting that $\frac{n}{n+2} \to 1$, it suffices to show that $(n+2)I_n$ converges to $\frac{1}{4}$. However, by integration by parts,

\begin{align*} (n+2) I_n &= \int_{0}^{1} (n+2)x^{n+1} \cdot \frac{1}{x+3} \, \mathrm{d}x \\ &= \underbrace{\left[ x^{n+2} \cdot \frac{1}{x+3} \right]_{0}^{1}}_{=\frac{1}{4}} + \int_{0}^{1} \frac{x^{n+2}}{(x+3)^2} \, \mathrm{d}x. \end{align*}

Moreover, from the inequality $0\leq \frac{1}{(x+3)^2} \leq \frac{1}{9}$ which holds true for $x \in [0, 1]$, we get

$$ 0 \leq \int_{0}^{1} \frac{x^{n+2}}{(x+3)^2} \, \mathrm{d}x \leq \int_{0}^{1} \frac{x^{n+2}}{9} \, \mathrm{d}x = \frac{1}{9(n+3)}. $$

Combining altogether, we get

$$ \frac{1}{4} \leq (n+2)I_n \leq \frac{1}{4} + \frac{1}{9(n+3)} $$

and therefore $(n+2) I_n \to \frac{1}{4}$ as required.

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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\lim_{n \to \infty}\pars{n\int_{0}^{1}{x^{n + 1} \over x + 3}} \,\dd x} = \lim_{n \to \infty}\bracks{n\int_{0}^{1}{\pars{1 - x}^{n + 1} \over \pars{1 - x} + 3}\,\dd x} \\[5mm] = &\ \lim_{n \to \infty}\bracks{n\int_{0}^{1} {\expo{\pars{n + 1}\ln\pars{1 - x}} \over 4 - x}\,\dd x} = \lim_{n \to \infty}\bracks{n\int_{0}^{\infty} {\expo{-\pars{n + 1}x} \over 4 - 0}\,\dd x} \\[5mm] = &\ {1 \over 4}\lim_{n \to \infty}\pars{n \over n + 1} = \bbx{\large{1 \over 4}} \\ & \end{align} See Laplace's Method.

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