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This is exercise from my lecturer, for IMC preparation. I haven't found any idea.

Find the value of

$$\lim_{n\rightarrow\infty}n^2\left(\int_0^1 \left(1+x^n\right)^\frac{1}{n} \, dx-1\right)$$

Thank you

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3 Answers 3

up vote 8 down vote accepted

By integration by parts,

\begin{align*} \int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx &= \left[ -(1-x)(1+x^{n})^{\frac{1}{n}} \right]_{0}^{1} + \int_{0}^{1} (1-x)(1 + x^{n})^{\frac{1}{n}-1}x^{n-1} \, dx \\ &= 1 + \int_{0}^{1} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx \end{align*}

so that we have

\begin{align*} n^{2} \left( \int_{0}^{1} (1 + x^{n})^{\frac{1}{n}} \, dx - 1 \right) &= \int_{0}^{1} n^{2} (1-x) (1 + x^{n})^{\frac{1}{n}-1} x^{n-1} \, dx. \end{align*}

Let $a_{n}$ denote this quantity. By the substitution $y = x^{n}$, it follows that

\begin{align*} a_{n} &= \int_{0}^{1} n \left(1-y^{1/n}\right) (1 + y)^{\frac{1}{n}-1} \, dy = \int_{0}^{1} \int_{y}^{1} t^{\frac{1}{n}-1} (1 + y)^{\frac{1}{n}-1} \, dtdy \end{align*}

Since $0 \leq t (1 + y) \leq 2$ and $ \int_{0}^{1} \int_{y}^{1} t^{-1}(1+y)^{-1} \, dtdy < \infty$, an obvious application of the dominated convergence theorem shows that

\begin{align*} \lim_{n\to\infty} a_{n} = \int_{0}^{1} \int_{y}^{1} \frac{dtdy}{t(1+y)} &= - \int_{0}^{1} \frac{\log y}{1+y} \, dy \\ &= \sum_{m=1}^{\infty} (-1)^{m} \int_{0}^{1} y^{m-1} \log y \, dy = \sum_{m=1}^{\infty} \frac{(-1)^{m-1}}{m^2} = \frac{\pi^2}{12}. \end{align*}

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Nice work! Now it's proven :) –  O.L. Jul 3 '13 at 10:29
    
Very nice! The value of $\pi^2/12 = .822...$ falls within my bounds of .75 to .875 which is comforting. –  marty cohen Jul 3 '13 at 13:33
    
@sos440 here is a shortcut I noticed. By DCT $ \lim_{n\to\infty}\int_{0}^{1} n \left(1-y^{1/n}\right) (1 + y)^{\frac{1}{n}-1} \, dy=\int_{0}^{1} \lim_{n\to\infty} \frac{\left(1-y^{1/n}\right)}{1/n} (1 + y)^{\frac{1}{n}-1} \, dy=-\int_{0}^{1}\frac{\log y}{1+y}\ dy=\frac{\pi^2}{12}$ –  Chris's sis Jul 3 '13 at 19:30
    
@Chris'swisesister, that's my first approach but I was unsure if we can find a dominating function for $(1 - y^{1/n})/(1/n)$. This is why I introduced double integral. –  sos440 Jul 4 '13 at 3:24
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I get an answer that differs from that of user17762. This is because the error term is not one term of order $\frac{x^{2n}}{n^2}$ but a number of such terms.

I get that the limit is between 3/4 and 7/8, but only have an infinite series for the value.

My complete analysis follows.

$(1+x^n)^{1/n} =\sum_{k=0}^{\infty} \binom{1/n}{k}x^{kn} $.

We first look at $\binom{1/n}{k}$.

$\begin{align} \binom{1/n}{k} &=\frac1{k!}\prod_{i=0}^{k-1}(\frac{1}{n}-i)\\ &=\frac1{k!n^k}\prod_{i=0}^{k-1}(1-in)\\ &=\frac{(-1)^k}{k!n^k}(-1)\prod_{i=1}^{k-1}(in-1)\\ &=\frac{(-1)^{k+1}}{k!n^k}\prod_{i=1}^{k-1}(in-1)\\ \end{align} $

so

$\begin{align} \big|\binom{1/n}{k}\big| =\frac1{k!n^k}\prod_{i=1}^{k-1}(in-1) <\frac{1}{k!n^k}\prod_{i=1}^{k-1}(in) =\frac{n^{k-1}(k-1)!}{k!n^k} =\frac1{kn} \end{align} $

and

$\begin{align} \frac{\binom{1/n}{k+1}}{\binom{1/n}{k}} &=\frac{(-1)^{k+2}}{(k+1)!n^{k+1}}\frac{k!n^k}{(-1)^{k+1}}\frac{\prod_{i=1}^{k}(in-1)}{\prod_{i=1}^{k-1}(in-1)}\\ &=\frac{-1}{(k+1)n}(kn+1)\\ &=-\frac{kn+1}{kn+n}\\ \end{align} $

We now look at $\int_0^v (1+x^n)^{1/n}\, dx$ to see what happens as $v \to 1$.

$\begin{align} \int_0^v (1+x^n)^{1/n}\, dx &=\sum_{k=0}^{\infty} \binom{1/n}{k} \int_0^v x^{kn}\, dx\\ &=\sum_{k=0}^{\infty} \binom{1/n}{k} \frac{v^{kn+1}}{kn+1}\\ &=v+\frac{v^{n+1}}{n(n+1)}+\sum_{k=2}^{\infty} \binom{1/n}{k} \frac{v^{kn+1}}{kn+1}\\ &=v+\frac{v^{n+1}}{n(n+1)}+v^{2n+1}\sum_{k=2}^{\infty} \binom{1/n}{k} \frac{v^{(k-2)n}}{kn+1}\\ \end{align} $

This means that the terms in $\sum_{k=2}^{\infty} \binom{1/n}{k} \frac{v^{(k-2)n}}{kn+1} $ decrease in absolute value and, since they alternate in sign, the series converges. and converges even at $v=1$ because of the $\frac1{kn+1}$.

Let $f(v, n) = \sum_{k=2}^{\infty} \binom{1/n}{k} \frac{v^{(k-2)n}}{kn+1} $.

Since $\big|\binom{1/n}{k} \frac{1}{kn+1}\big| < \frac1{kn(kn+1)} $, $|f(v, n)| <\sum_{k=2}^{\infty} \frac1{kn(kn+1)} <\frac1{n^2}\sum_{k=2}^{\infty} \frac1{k^2} <\frac1{n^2} $.

The first term of $f(v, n)$ is $\binom{1/n}{2}\frac{1}{2n+1} =\frac{(1/n)(1/n-1)}{2}\frac{1}{2n+1} =-\frac{n-1}{2n^2(2n+1)} $ and this is between $-\frac1{8n^2}$ and $-\frac1{4n^2}$ for $n > 3$.

Therefore the first two terms of the expansion of $\int_0^v (1+x^n)^{1/n}\, dx$ are both of order $1/n^2$, so we have to consider the whole sum, not just the first term (after $1$).

Since $\int_0^v (1+x^n)^{1/n}\, dx =v+\frac{v^{n+1}}{n(n+1)}+v^{2n+1}f(v, n) $ and all the terms exist as $v \to 1$, $\int_0^1 (1+x^n)^{1/n}\, dx -1-\frac{1}{n(n+1)}=f(1, n) $.

Since $f(1,n)$ is between $-\frac1{8n^2}$ and $-\frac1{4n^2}$, $-\frac1{4n^2} <\int_0^1 (1+x^n)^{1/n}\, dx -1-\frac{1}{n(n+1)} <-\frac1{8n^2} $, $-\frac1{4} <n^2\big(\int_0^1 (1+x^n)^{1/n}\, dx -1\big)-\frac{n}{n+1} <-\frac1{8} $ so $1-\frac1{4} <\lim_{n \to \infty} n^2 \big(\int_0^1 (1+x^n)^{1/n}\, dx -1\big) < 1-\frac1{8} $.

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+1. Actually, I have found that the exact value is equal to $\pi^2/12$. –  O.L. Jul 3 '13 at 7:58
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Mathematica evaluates the integral to $$\int_0^{1}(1+x^n)^{1/n}dx={}_2F_1\left(-\frac{1}{n},\frac1n,1+\frac1n;-1\right).\tag{1}$$ Next, let us write the standard series representation for the hypergeometric function $$_2F_1(a,b,c;t)=\sum_{k=0}^{\infty}\alpha_kt^k,\qquad \alpha_k=\frac{\Gamma(a+k)\Gamma(b+k)\Gamma(c)}{k!\,\Gamma(a)\Gamma(b)\Gamma(c+k)}.$$ Now an easily verified claim: as $n\rightarrow\infty$, for the parameters as in (1), we have $\alpha_0=1$ and $$\alpha_k\sim-\frac{1}{k^2n^2}+O(n^{-3}).$$ Hence we obtain \begin{align} \lim_{n\rightarrow\infty}n^2\left(\int_0^{1}(1+x^n)^{1/n}dx-1\right)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2}=\frac{\pi^2}{12}, \end{align} which is also confirmed by numerical calculation with $n\sim 10^4-10^8$.

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