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I was trying to compute $\lim\limits_{n\rightarrow\infty} \int_{0}^{1} \frac{e^{-nt}-(1-t)^n}{t} dt$ using Lebesgue's dominated convergence THM, but I can't exactly figure out how to do. I mean, I managed to prove that each the integrand function $f_n(t)$ is less or equal than $g_n(t)=e^{-nt}\sqrt{n}e^\frac{1}{\sqrt{n}}\, \forall n\in\mathbb{N}$. And since we are dealing with positive functions and $\int_{0}^{1} g_n(t)dt\leq\frac{e}{\sqrt{n}}\overset{\mathrm{n\rightarrow\infty}}{\rightarrow}0$, I can deduce that the original limit is $0$.


Now, I was just wondering if anyone is able to show analytically that there exists a function in $\mathcal{L}^1$ which dominates all the $f_n$ in order to apply Lebesgue's dominated convergence THM. I made some attempts, but I failed.


Thanks in advance, a humble half-mathematician.

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We have $$\exp(-nt)-(1-t)^n=\exp(-nt)\,\Big(1-\big((1-t)\exp(t)\big)^n\Big)\,.\tag{*}$$ By Bernoulli's Inequaliy, $$\big((1-t)\exp(t)\big)^n\geq 1+n\,\big((1-t)\exp(t)-1\big)\,.$$ Therefore, $$\exp(-nt)-(1-t)^n\leq n\,\exp(-nt)\,\big(1-(1-t)\exp(t)\big)\,.$$ By taking derivative with respect to $n$, we can show that $$n\,\exp(-nt)\leq \frac{1}{\text{e}\,t}\text{ for all }t>0\text{ and positive integers }n\,.$$ Furthermore, we have $$(1-t)\,\exp(t)=1-\sum_{k=1}^\infty\,\left(\frac{k-1}{k!}\right)\,t^k\geq 1-t^2\,\sum_{k=2}^\infty\,\frac{k-1}{k!}=1-t^2$$ for all $t\in[0,1]$.

By (*), we conclude that $$f_n(t):=\frac{\exp(-nt)-(1-t)^n}{t}\leq \frac{1}{\text{e}\,t^2}\,\big(1-(1-t^2)\big)=\frac{1}{\text{e}}$$ for every $t\in (0,1]$ and every positive integer $n$ (where the only equality case is when $n=1$ and $t=1$). Therefore, the constant function $f\equiv \dfrac{1}{\text{e}}$ dominates $f_n$ for all $n=1,2,3,\ldots$.

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  • $\begingroup$ Absolutely brilliant. Thanks a lot. $\endgroup$ – piotor Dec 8 '18 at 18:45

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