# Real Analysis, $\lim\limits_{n\rightarrow\infty} \int_{0}^{1} \frac{e^{-nt}-(1-t)^n}{t} dt$

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.

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 $$n=1,2,3,\ldots$$.