Compute Limit: \sqrt[n]{1+x^n}^n

[MarkFL]

Compute the following limit:

$$\lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)$$

 

[Joppy]

Compute the following limit:

$$\lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)$$

Saw this on another site. Is there a bounty out for it xD.

 

[Opalg]

Compute the following limit:

$$\lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)$$

[sp]For $0\leqslant x\leqslant1$, $1+x^n \leqslant2$. Therefore $$\int_0^1 \left(1+x^n\right)^n\,dx \leqslant 2^n$$ and hence $$\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx} \leqslant2.$$

Now choose $n$ and let $x_0 = \left(1-\frac1n \right)^{1/n}$. The function $\left(1+x^n\right)^n$ is increasing on $[0,1]$, so on the interval $[x_0,1]$ it is greater than or equal to $\left(1+x_0^n\right)^n = \left(2 - \frac1n\right)^n$. Therefore $$\int_0^1 \left(1+x^n\right)^n\,dx \geqslant \int_{x_0}^1 \left(1+x^n\right)^n\,dx \geqslant \int_{x_0}^1 \left(2 - \tfrac1n\right)^ndx = \left(1 - \left(1-\tfrac1n\right)^{1/n}\right) \left(2 - \tfrac1n\right)^n.$$ Thus $$2 \geqslant \sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx} \geqslant \left(1 - \left(1-\tfrac1n\right)^{1/n}\right)^{1/n} \left(2 - \tfrac1n\right).$$

The aim now is to show that $$\lim_{n\to\infty} \left(1 - \left(1-\tfrac1n\right)^{1/n}\right)^{1/n} = 1.$$ That will show that $$\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx} $$ can be made as close as we wish to $2$ for all sufficiently large $n$, and hence $$\lim_{n\to\infty}\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx} = 2.$$

Start with the estimates $-2x <\ln(1-x) < -x$ and $\frac x2 < 1-e^{-x} < x$, valid for $0<x<\frac12.$ Put $x=\frac1n$ in the first inequality to get $$-\frac2n < \ln\left(1-\frac1n\right) < -\frac1n$$ for all $n\geqslant 2.$ Then $$e^{-2/n^2} < \left(1-\tfrac1n\right)^{1/n} < e^{-1/n^2},$$ and $$\frac2{n^2} > 1 - e^{-2/n^2} > 1 - \left(1-\tfrac1n\right)^{1/n} > 1 - e^{-1/n^2} > \frac1{2n^2}.$$ Therefore $$\left(\frac2{n^2}\right)^{1/n} > \left(1 - \left(1-\tfrac1n\right)^{1/n}\right)^{1/n} > \left(\frac1{2n^2}\right)^{1/n}$$, and the result follows from the fact that $$\lim_{n\to\infty}\left(\frac1n\right)^{1/n} = 1.$$

[/sp]

 

[MarkFL]

[sp]For $0\leqslant x\leqslant1$, $1+x^n \leqslant2$. Therefore $$\int_0^1 \left(1+x^n\right)^n\,dx \leqslant 2^n$$ and hence $$\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx} \leqslant2.$$

Now choose $n$ and let $x_0 = \left(1-\frac1n \right)^{1/n}$. The function $\left(1+x^n\right)^n$ is increasing on $[0,1]$, so on the interval $[x_0,1]$ it is greater than or equal to $\left(1+x_0^n\right)^n = \left(2 - \frac1n\right)^n$. Therefore $$\int_0^1 \left(1+x^n\right)^n\,dx \geqslant \int_{x_0}^1 \left(1+x^n\right)^n\,dx \geqslant \int_{x_0}^1 \left(2 - \tfrac1n\right)^ndx = \left(1 - \left(1-\tfrac1n\right)^{1/n}\right) \left(2 - \tfrac1n\right)^n.$$ Thus $$2 \geqslant \sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx} \geqslant \left(1 - \left(1-\tfrac1n\right)^{1/n}\right)^{1/n} \left(2 - \tfrac1n\right).$$

The aim now is to show that $$\lim_{n\to\infty} \left(1 - \left(1-\tfrac1n\right)^{1/n}\right)^{1/n} = 1.$$ That will show that $$\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx} $$ can be made as close as we wish to $2$ for all sufficiently large $n$, and hence $$\lim_{n\to\infty}\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx} = 2.$$

Start with the estimates $-2x <\ln(1-x) < -x$ and $\frac x2 < 1-e^{-x} < x$, valid for $0<x<\frac12.$ Put $x=\frac1n$ in the first inequality to get $$-\frac2n < \ln\left(1-\frac1n\right) < -\frac1n$$ for all $n\geqslant 2.$ Then $$e^{-2/n^2} < \left(1-\tfrac1n\right)^{1/n} < e^{-1/n^2},$$ and $$\frac2{n^2} > 1 - e^{-2/n^2} > 1 - \left(1-\tfrac1n\right)^{1/n} > 1 - e^{-1/n^2} > \frac1{2n^2}.$$ Therefore $$\left(\frac2{n^2}\right)^{1/n} > \left(1 - \left(1-\tfrac1n\right)^{1/n}\right)^{1/n} > \left(\frac1{2n^2}\right)^{1/n}$$, and the result follows from the fact that $$\lim_{n\to\infty}\left(\frac1n\right)^{1/n} = 1.$$

[/sp]

Nicely done, Chris! (Yes)

This is the solution I found elsewhere:

Spoiler

We are given to evaluate:

$$\lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)$$

Let $$f(x)=1+x^n$$ and observe that for all $$x\in\left[0,1\right]$$ and $$n\in\mathbb{N}$$, we have $$0<f(x)$$.

Now, let's define:

$$M\equiv\sup_{x \in [0,1]} f(x)$$

Observe then that we must have:

$$\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}} \le \left(\int_0^1 M^n\,dx\right)^{\frac{1}{n}}= M$$

Thus, we conclude:

$$\limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M$$

Now let $\alpha$ be any non-negative real number strictly less than $M$. By definition, there must be some $x_0\in[0,1]$ such that $f\left(x_0\right)=M$. By continuity of $f$, we can find an interval $(c,d) \subset [0,1]$ such that $f(x)>\alpha$ for all $x\in(c,d)$. Then, for every $n$, we have:

$$\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\ge\left(\int_c^d f(x)^n\,dx\right)^{\frac{1}{n}}\ge\left(\int_c^d \alpha^n\,dx\right)^{\frac{1}{n}}=\alpha(d-c)^{\frac{1}{n}}$$

Taking limits, there results:

$$\liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\ge\lim_{n\to\infty}\left(\alpha(d-c)^{\frac{1}{n}}\right)=\alpha$$

Given that $\alpha$ is an arbitrary real number strictly less than $M$, the above implies:

$$\liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\ge M$$

Thus, we have:

$$M\le\liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le\limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M$$

And this implies:

$$\lim_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)=M$$

For $f(x)=1+x^n$, we find $M=2$.

 

[Opalg]

This is the solution I found elsewhere:

Spoiler

We are given to evaluate:

$$\lim_{n\to\infty}\left(\sqrt[n]{\int_0^1 \left(1+x^n\right)^n\,dx}\right)$$

Let $$f(x)=1+x^n$$ and observe that for all $$x\in\left[0,1\right]$$ and $$n\in\mathbb{N}$$, we have $$0<f(x)$$.

Now, let's define:

$$M\equiv\sup_{x \in [0,1]} f(x)$$

Observe then that we must have:

$$\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}} \le \left(\int_0^1 M^n\,dx\right)^{\frac{1}{n}}= M$$

Thus, we conclude:

$$\limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M$$

Now let $\alpha$ be any non-negative real number strictly less than $M$. By definition, there must be some $x_0\in[0,1]$ such that $f\left(x_0\right)=M$. By continuity of $f$, we can find an interval $(c,d) \subset [0,1]$ such that $f(x)>\alpha$ for all $x\in(c,d)$. Then, for every $n$, we have:

$$\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\ge\left(\int_c^d f(x)^n\,dx\right)^{\frac{1}{n}}\ge\left(\int_c^d \alpha^n\,dx\right)^{\frac{1}{n}}=\alpha(d-c)^{\frac{1}{n}}$$

Taking limits, there results:

$$\liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\ge\lim_{n\to\infty}\left(\alpha(d-c)^{\frac{1}{n}}\right)=\alpha$$

Given that $\alpha$ is an arbitrary real number strictly less than $M$, the above implies:

$$\liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\ge M$$

Thus, we have:

$$M\le\liminf_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le\limsup_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)\le M$$

And this implies:

$$\lim_{n\to\infty}\left(\left(\int_0^1 f(x)^n\,dx\right)^{\frac{1}{n}}\right)=M$$

For $f(x)=1+x^n$, we find $M=2$.

[sp]That argument does not work, because the function $f(x) = 1 + x^n$ depends on $n$. In fact, there is a whole sequence of functions $f_n(x) = 1 + x^n$. As it happens, they all have the same supremum $M = 2$. Given $\alpha<M$, each of these functions will have an interval $(c,d)$ on which $f_n(x)>\alpha$. But that interval will depend on $n$, so it should really be written $(c_n,d_n)$. As $n$ increases, that interval may well get shorter and shorter, and no reason is given to justify the conclusion that $(d_n - c_n)^{1/n} \to 1$ as $n\to\infty$. A more delicate argument is needed for that.

In my proof, I used $(x_0,1)$ as the interval on which $f_n(x) > 2 - \frac1n$. There too, I was using a possibly misleading notation, because $x_0$ also depends on $n$. I should probably have called it $x_n$. But my proof takes account of the fact that the interval $(x_n,1)$ decreases as $n$ increases.

So I think that my solution is right and the other one is wrong. :p

[/sp]