This question is from Bass [Real Analysis for Graduate Students], Exercise 11 of chapter 7, which deals with monotone convergence theorem(MCT), Fatou's lemma, Lebesgue Dominated Convergence Theorem (LDCT).

Find the limit $$\lim_{n\to\infty} \int_0^n(1+ \frac xn)^{-{n}} \log(2+\cos(x/n))dx$$ and justify your reasoning.

I guess the limit equals to $\displaystyle\lim_{n\to\infty}\int_0^{\infty} e^{-x}\log3dx$, by applying the pointwise convergence, but I am stuck how to justify my guess. I'd like to apply LDCT, but how should I do for $n$ of the integral?

I first tried to apply MCT, but the sequence of functions $f_n (x) = (1+\frac xn )^{-n}$ decreases as $n$ becomes large, so it failed.

Next I tried to apply LDCT after changing variables as $t= \frac xn$, so the integral becomes $\lim_{n \to\infty}\int_0^1(1+t)^{-n}\log(2+\cos t)ndt$. However, the maximum value of the integrand goes to infinity as $n \to \infty$, so it also failed.

Does anyone have any idea?


1 Answer 1


Note that for $0 \leqslant x \leqslant n$ we have

$$\left(1 + \frac{x}{n} \right)^{-n} \log(2 + \cos(x/n))\leqslant (\log 3)e^{-x/2},$$

since, using the inequality $\log(1+y) \geqslant y/(1+y)$, we have

$$n \log(1 + x/n) \geqslant n \frac{x/n}{1 + x/n} \geqslant \frac{x}{2} \\ \implies - n \log(1 + x/n) \leqslant - \frac{x}{2} \\ \implies \left(1 + \frac{x}{n} \right)^{-n} \leqslant e^{-x/2} $$

You can now apply LDCT to

$$\int_0^n \left(1 + \frac{x}{n} \right)^{-n} \log(2 + \cos(x/n)) \, dx = \int_0^\infty \left(1 + \frac{x}{n} \right)^{-n} \log(2 + \cos(x/n)) \chi_{[0,n]}(x) \, dx $$

  • $\begingroup$ How should I explain that $\int_0^n$ goes to $\int_0^{\infty}$? MCT only holds when the interval of integration is the same. $\endgroup$
    – bellcircle
    Commented Feb 2, 2017 at 3:43
  • $\begingroup$ I revised the exponent as $(1 \pm x/n)^{\mp n}$. $\endgroup$
    – bellcircle
    Commented Feb 2, 2017 at 3:47
  • $\begingroup$ If the exponent is $-n$, instead of $n$, then I can't apply Bernoulli's inequality. Then how should I do? $\endgroup$
    – bellcircle
    Commented Feb 2, 2017 at 3:52
  • $\begingroup$ Sorry, I found the typo right ago. $\endgroup$
    – bellcircle
    Commented Feb 2, 2017 at 3:53
  • $\begingroup$ @bellcircle: One more try $\endgroup$
    – RRL
    Commented Feb 3, 2017 at 3:52

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