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Establish the following limits:

  1. $\lim_{n\to \infty}\int_{[0,1]}\dfrac{n^2x^2}{\exp(nx)}=0$
  2. $\lim_{n\to \infty}\int_{[0,1]}\dfrac{1+nx}{(1+x)^n}=0$
  3. $\lim_{n\to \infty}\int_{[0,\infty]}\left(1+\dfrac{x}{n}\right)^n\exp(-ax)=\dfrac{1}{a-1}$ (In the last part, assume $a>1$.)

The first two were relatively simple to do. Just do tabular integration for each and then take the limit and you are done.

However the third one is giving me a hard time to show that the statement is true. I first integrated and found this large complex expansion. But this seems too complicated.

Can I get some suggestions?

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  • $\begingroup$ Did you know that $\exp(c):=\lim_{n\rightarrow \infty} (1+c/n)^n$ and that further, this sequence is increasing? $\endgroup$ Commented Apr 6, 2016 at 21:50
  • $\begingroup$ Yes. So I should move the lim inside the integral? $\endgroup$ Commented Apr 6, 2016 at 21:51
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    $\begingroup$ If you know the MCT, you may switch limit and integration. $\endgroup$
    – Hetebrij
    Commented Apr 6, 2016 at 21:54
  • $\begingroup$ Yes, you can justify this using the squeeze theorem. MCT is a good way too $\endgroup$ Commented Apr 6, 2016 at 21:54
  • $\begingroup$ So I have done the switching for the first two problems? $\endgroup$ Commented Apr 6, 2016 at 21:55

2 Answers 2

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Note that

$$\left(1+\frac xn\right)^n\le e^x$$

for $x>0$ and $n\ge 0$. Therefore, the Dominated Convergence Theorem guarantees that we have

$$\lim_{n\to \infty}\int_0^\infty \left(1+\frac xn\right)^n\,e^{-ax}\,dx=\int_0^\infty \,e^{-(a-1)x}\,dx=\frac{1}{a-1}$$

for $a>1$.

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Using Bernoulli's Inequality, $$ \begin{align} \frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}n\right)^n} &=\left(\frac{(n+x+1)n}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right)\\ &=\left(1-\frac x{(n+x)(n+1)}\right)^{n+1}\frac{n+x}n\\[6pt] &\ge\left(1-\frac x{n+x}\right)\frac{n+x}n\\[12pt] &=1 \end{align} $$ That is, $\left(1+\frac xn\right)^n$ is increasing in $n$.

Thus, by Monotone Convergence, $$ \begin{align} \lim_{n\to\infty}\int_0^\infty\left(1+\frac xn\right)^n\,e^{-ax}\,\mathrm{d}x &=\int_0^\infty\lim_{n\to\infty}\left(1+\frac xn\right)^n\,e^{-ax}\,\mathrm{d}x\\ &=\int_0^\infty e^x\,e^{-ax}\,\mathrm{d}x\\[3pt] &=\frac1{a-1} \end{align} $$

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