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?


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
    Feb 2 '17 at 3:43
  • $\begingroup$ I revised the exponent as $(1 \pm x/n)^{\mp n}$. $\endgroup$
    – bellcircle
    Feb 2 '17 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
    Feb 2 '17 at 3:52
  • $\begingroup$ Sorry, I found the typo right ago. $\endgroup$
    – bellcircle
    Feb 2 '17 at 3:53
  • $\begingroup$ @bellcircle: One more try $\endgroup$
    – RRL
    Feb 3 '17 at 3:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.