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Find $$\lim_{n \rightarrow \infty} \int_0 ^{\infty} \left(1+ \frac{x}{n}\right)^n e^{-ex} dx.$$

I know that $\lim_{n \rightarrow \infty} \left(1+ \frac{x}{n}\right)^n = e^x.$

Also, I showed that $ \left(1+ \frac{x}{n}\right)^n \leq e^x$ for all $x \geq -n$.

So I'm thinking about the Lebesgue Dominated Convergence Theorem. But I'm not sure how to deal with the $e^{-ex}$ part.

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    $\begingroup$ How about $g(x) := e^{x }e^{-ex} = e^{(1 - e)x}$ being the dominating function? Since $1 - e < 0$, the integral of $g$ does exist. $\endgroup$ Commented Feb 8, 2021 at 3:54

3 Answers 3

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Let $f_n : [0, \infty) \to \Bbb R$ be defined by $$f_n(x) := \left(1 + \frac{x}{n}\right)^ne^{-ex}.$$

With a clever use of AM-GM, one can show that $f_n(x) \le f_{n+1}(x)$ for all $x \ge 0$. (You only need to show the inequality for the part without the $e^{-ex}$.)

What can you conclude using the Monotone Convergence Theorem now?

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$$\int \left(1+ \frac{x}{n}\right)^n e^{-ex}\, dx= {n^{-n}}\int(x+n)^ne^{-ex}\, dx$$ $$\int (x+a)^b e^{-c x}\, dx=-e^{a c} c^{-b-1} \Gamma (b+1,c (x+a))$$ $$\int_0^\infty (x+a)^b e^{-c x}\, dx=e^{a c} c^{-b-1} \Gamma (b+1,a c)\quad \text{if}\quad \Re(c)>0\land \Re(a)>0$$ All of that makes $$I_n=\int_0^\infty \left(1+ \frac{x}{n}\right)^n e^{-ex}\, dx=e^{(e-1) n-1}\, n^{-n} \,\Gamma (n+1,e n)$$

Have a look here [at $§8.11(iii)$] for $\lambda=e$.

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Let $$ f_n(x)=\left(1 + \frac{x}{n}\right)^ne^{-ex} \chi_{[0,n]}(x).$$ Then $$ |f_n(x)|\le e^{-(e-1)x}, \lim_{n\to\infty}f_n(x)=f(x)=e^{-(e-1)x}. $$ Since $$ \int_0^\infty e^{-(e-1)x}dx<\infty $$ the DCT, one has $$ \lim_{n\to\infty}\int_0^\infty f_n(x)dx=\int_0^\infty\lim_{n\to\infty} f_n(x)dx. $$

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