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I considered the series $I_{n}=\int_{0}^{n}\sqrt[n]{x}\cdot e^{-x}dx$ und while calculating $\lim_{n\rightarrow \infty}I_n$ I wasn't sure how to use the dominated converge theorem which allows me to exchange limit and integration. I verified the conditions to use it and I know how to calculate, but the problem is the correct notation because $n$ is also a part of the bounds of integration.

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    I don't know measure/integration theory too well, but I know one trick is to extend this function to all of $[0,\infty)$ by taking the function to be zero outside $[0,n]$. Maybe this will work? – 2017-02-01

2 Answers 2

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Hint. One may write $$ \int_{0}^{n}\sqrt[n]{x}\cdot e^{-x}dx=\int_{0}^{\infty}1_{[0,n]}(x)\cdot\sqrt[n]{x}\cdot e^{-x}dx $$ with $1_{[0,n]}(x)=1$ if $x \in [0,n]$ and $1_{[0,n]}(x)=0$ if $x \notin [0,n]$.

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note that $\sqrt[n]{x}=e^{\frac{1}{n}ln\ x}\leq e^{ln\ x}=x$, if $x\geq 1$. So, $\mathbb{1_{[0,n]}}(x)\sqrt[n]{x}e^{-x}\leq \mathbb{1}_{[0,1]}(x)(1+xe^{-x})$. Since $\mathbb{1}_{[0,1]}(x)(1+xe^{-x})$ is integrable, by the dominated converge theorem we have: $$ \mathbb{lim}_{n\rightarrow \infty}\int_{0}^{n}\sqrt[n]{x}e^{-x}=\mathbb{lim}_{n\rightarrow \infty}\int_{\mathbb{R}}\mathbb{1_{[0,n]}}(x)\sqrt[n]{x}e^{-x}=\int_{\mathbb{R}}\mathbb{lim}_{n\rightarrow \infty}\mathbb{1_{[0,n]}}(x)\sqrt[n]{x}e^{-x}=\int_{\mathbb{R}}\mathbb{1_{[0,\infty]}}(x)e^{-x} $$

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    It is not correct that $\sqrt[n]{x}\leq x$ in $[0,1]$, but that can be fixed. – 2017-02-01
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    How can I fix it? – 2017-02-01
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    you're right. fixed – 2017-02-01
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    Another problem: $xe^{-x}< 0$ if $x<0$ but in my opinion we need a positive majoring. Am I wrong? – 2017-02-01
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    Sorry but the upper bound $$\mathbb{1_{[0,n]}}(x)\sqrt[n]{x}e^{-x}\leq \mathbb{1}_{[0,1]}(x)(1+xe^{-x})$$ is (obviously) wrong **for every $n\geqslant2$**. (Why the hasty acceptance of this answer?) – 2017-02-02
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    Would be $f(x)=2$ a majoring to fix this problem? – 2017-02-04