Show that $$\lim_{n\rightarrow\infty}\int_{0}^{\infty}\frac{x^{\frac{1}{n}}}{(1+\frac{x}{n})^{n}}dx=1.$$

Remark: I must accept that I have a conflict when I am faced with a situation where I must know the relationship between Integrals of Riemann and Integrals of Lebesgue, the conflict is stronger when integrals are improper. Most books address this relationship for integrals over intervals of the form $ [a, b] $, but very few address improper integrals. I would like to know the proof of the problem that I have proposed and that someone recommend me books where they propose exercises similar to the one I have proposed here.


Notice that the pointwise limit of the integrands is $e^{-x}$. Moreover, notice that for $M \ge 1$ large enough $$f_n(x) = \frac{x^{\frac{1}{n}}}{(1 + \frac{x}{n})^n} \le \frac{1}{x^2}.$$ Then $$g(x) = \begin{cases}M & \text{if}\ x\in (0,M)\\ \frac{1}{x^2} & \text{if}\ x\in [M,\infty) \end{cases}$$ is integrable and satisfies $|f_n| \le g$ in $(0,\infty)$. By Lebesgue's Dominated Convergence Theorem, $$\lim_n\int_0^{\infty}f_n\,dx = \int_0^{\infty}\lim_nf_n\,dx = \int_0^{\infty}e^{-x}\,dx = 1.$$

Every book on measure theory covers this theorem and its applications. As a good starting point I'd suggest Real Analysis: Modern Techniques and Their Applications by Gerald B. Folland.

  • $\begingroup$ (+1) for the well-constructed solution. Since the binomial theorem provides such a clean bound, this is the way I would have proceeded. $\endgroup$ – Mark Viola May 5 '17 at 4:16
  • $\begingroup$ @Dr.MV: I have +1 your answer as well. I wish I could say that I would have proceeded as you did but the truth is that I have absolutely no experience with the kind of computations you performed. Interesting approach, thanks for sharing it :) $\endgroup$ – Giovanni May 5 '17 at 4:23
  • $\begingroup$ Thank you my friend! Much appreciative and pleased that you found my solution worthwhile. -Mark $\endgroup$ – Mark Viola May 5 '17 at 4:25

While the following solution does not rely on use of the Dominated Convergence Theorem (@Giovanni already provided that approach), I thought it might be instructive to present a way forward that relies on elementary analysis of the Beta and Gamma Functions. It is to this end that we now proceed.

Let $I(n)$ be given by

$$I(n)=\int_0^\infty \frac{x^{1/n}}{\left(1+\frac xn\right)^n}\,dx\tag 1$$

Enforcing the substitution $x\to nx$ in $(1)$ reveals

$$\begin{align} I(n)&=n^{1+1/n}\int_0^\infty \frac{x^{1/n}}{(1+x)^n}\,dx\\\\ &=n^{1+1/n}B\left(1+\frac1n,n-1-\frac1n\right)\\\\ &=n^{1+1/n}\frac{\Gamma\left(1+\frac1n\right)\Gamma\left(n-1-\frac1n\right)}{\Gamma\left(n\right)} \end{align}$$

Noting that as $n\to \infty$, $\Gamma\left(n-1-\frac1n\right)\sim \Gamma(n-1)=(n-2)!$, $\Gamma\left(1+\frac1n\right)\sim 1$, we find that

$$\begin{align} n^{1+1/n}\frac{\Gamma\left(1+\frac1n\right)\Gamma\left(n-1-\frac1n\right)}{\Gamma\left(n\right)}&\sim \frac{n (n-2)!}{(n-1)!}\\\\ &\to 1\,\,\text{as}\,\,n\to \infty \end{align}$$

  • $\begingroup$ I believe there's a typo in the very last line :) $\endgroup$ – Giovanni May 5 '17 at 4:14
  • 1
    $\begingroup$ @Giovanni Yes, there sure was. Thank you for the nice catch! I've edited. -Mark $\endgroup$ – Mark Viola May 5 '17 at 4:15
  • $\begingroup$ Shouldn't the last line say "goes to 1", not "0"? $\endgroup$ – bloomers May 5 '17 at 4:16
  • $\begingroup$ @bloomers It does; have a look. $\endgroup$ – Mark Viola May 5 '17 at 4:17

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