It is required to calculate $\lim_{n\to\infty}\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx$. The following is my attempt.

$\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx=\int_{[0,\infty)}(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$ for each $n\in\mathbb{N}$. Let $g_n(x)=(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$ for each $n\in\mathbb{N}$ and $x\in[0,\infty) $.Then $g_n(x)\leq e^{-x}$ for each $x\in[0,\infty)$ and $n\in\mathbb{N}$. $\int_{[0,\infty)}e^{-x}dx$ exists. Moreover $g_n(x)$ converges to $g(x)=e^{-x}$ for each $x$. Then by Dominated Convergence theorem $\lim_{n\to\infty}\int_{[0,n]}(1+\frac{x}{n})^ne^{-2x}dx=\int_{[0,\infty)}e^{-x}dx$.

Could someone please tell me if my solution is alright? Thanks.

  • 1
    $\begingroup$ That is absolutely fine. $\endgroup$ May 18, 2017 at 3:45
  • 1
    $\begingroup$ You're reasoning is solid. $\endgroup$
    – Mark Viola
    May 18, 2017 at 4:17
  • 1
    $\begingroup$ Thank you very much Shark and Mark :) $\endgroup$
    – Janitha357
    May 18, 2017 at 7:52

1 Answer 1


The reasoning is correct. There is only one small typo in the definition of $g_n$: it should be $g_n(x)=(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}$ instead of $g_n(x)=(1+\frac{x}{n})^ne^{-2x}\chi_{[0,n]}dx$.


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