Finding $\lim_{n\to \infty}\int_{(0,\infty)}\frac{dt}{\left(1+\frac{t}{n}\right)^nt^{1/n}}$

Question

Q) Find $$\lim_{n\to \infty}\int_{(0,\infty)}\frac{dt}{\left(1+\frac{t}{n}\right)^nt^{1/n}}$$

I was hoping to apply Dominated convergence theorem to use $$\lim_{n\to \infty}\left(1+\frac{t}{n}\right)^{-n}= e^{-t}$$

but clearly I know the lower bounds but not an upper bound function which is absolutely integrable.

$$\left(1+\frac{t}{n}\right)^{-n}\geq e^{-t},t^{-1/n}\geq t^{-1} \quad\text{at least on } (1,\infty)$$

Can I find an upper bound or have to figure out the integral itself first? Thanks.

Answer 1

On $(0,1]$, the sequence $\{t^{-1/n}\}_{n=1}^{\infty}$ is decreasing. And $\{(1+t/n)^{n}\}_{n=1}^{\infty}$ is increasing for all $t>0$, and $\{(1+t/n)^{-n}\}_{n=1}^{\infty}$ is decreasing.

Note that $\displaystyle\int_{0}^{1}\dfrac{1}{(1+t/2)^{2}}\dfrac{1}{t^{1/2}}dt\leq\int_{0}^{1}\dfrac{1}{t^{1/2}}dt<\infty$, so by Monotone Convergence Theorem $\displaystyle\int_{0}^{1}\dfrac{1}{(1+t/n)^{n}}\dfrac{1}{t^{1/n}}dt\rightarrow\int_{0}^{1}e^{-t}dt$.

Note that Monotone Convergence Theorem has the decreasing version, as long as $f_{1}\in L^{1}$, $f_{1}\geq f_{2}\geq\cdots$, then the integrals correspond to them also converge.

Now we look at the interval $[1,\infty)$. We have \begin{align*} \dfrac{1}{(1+t/n)^{n}}\dfrac{1}{t^{1/n}}\leq\dfrac{1}{(1+t/n)^{n}}\leq\dfrac{1}{(1+t/2)^{2}} \end{align*} and $\displaystyle\int_{1}^{\infty}\dfrac{1}{(1+t/2)^{2}}dt<\infty$, an integrable upper bound is obtained.

Answer 2

Denote $f_n=\frac{1}{\left(1+\frac{t}{n}\right)^nt^{1/n}}$

Then $f_n(t) \to e^{-t}$ on $(0,\infty)$

Also $\forall n \geq 2$ $$(1+\frac{t}{n})^n=\sum_{k=0}^n \binom{n}{k}(\frac{t}{n})^k\geq 1+\frac{t^2}{2}$$

So $$\int_0^{\infty}|f_n| \leq \int_0^{\infty}\frac{1}{1+\frac{t^2}{2}}\frac{1}{\sqrt[n]{t}}dt = \int_0^1\frac{1}{1+\frac{t^2}{2}}\frac{1}{\sqrt[n]{t}}dt+\int_1^{\infty}\frac{1}{1+\frac{t^2}{2}}\frac{1}{\sqrt[n]{t}}dt=I_1+I_2$$

$I_2 \leq \int_1^{\infty}\frac{1}{t^2/2+1}dt<+\infty$

$I_1 \leq \int_0^1 \frac{1}{\sqrt{t}}dt$

Thus $|f_n| \leq \frac{1}{t^2/2+1}1_{[1,+\infty]}+\frac{1}{\sqrt{t}}1_{(0,1)} \in L^1((0,\infty)$

Use D.C.T