Use the Lebesgue's Monotone Convergence Theorem and the fact that $\int^\infty_0t^ne^{-t}dt=n! \text{ to prove that }\int^\infty_0f(t)e^{-t}dt=s.$

Question

Consider a sequence of real numbers $(a_n$ with $n\in\mathbb{N}_0$, such that $a)n \geq0\text { non-negative}$ for all $n\in\mathbb{N}_0$ and $\sum^\infty_{n=0 } =s\in\mathbb{R}$

b. Use the Lebesgue's Monotone Convergence Theorem and the fact that $\int^\infty_0t^ne^{-t}dt=n! \text{ to prove that }\int^\infty_0f(t)e^{-t}dt=s.$

From part a. $f(t):=\sum^\infty_0 a_n\frac{t^n}{n!}$ where $t\in\mathbb{R}$

My problem is that I don't see how the Theorem or the fact given relates. I know that $f(t)$ and $e^{-t}$ converge on range $0,\infty$ and that seems like good enough reason to say that the equation is true.

Lebesgue's Monotone Convergence Theorem based on my book (I didn't write the assumptions, just the results):

1. $\forall n\in\mathbb{N},\int_Af_n(x)dx\leq\int_0f_{n+1}(x)dx\leq\int_Af(x)dx$

2.$\lim_{n\rightarrow\infty}\int_Af_n(x)dx=\int_A\lim_{n\rightarrow\infty}f_n(x)dx=\int_Af(x)dx$

Answer 1

Suggestion: $f_n(t) = \sum_{k=0}^{n} a_k\frac{t^k}{k!}$.

Use the given fact to derive a formula for $\int_0^{\infty} f_n(t) e^{-t} \, dt$. Use the Monotone Convergence Theorem to justify $\lim \int f_n(t) e^{-t} \, dt = \int \lim f_n(t) e^{-t} \, dt$.

Answer 2

Since we know that $f(t) = \sum_{n = 0}^\infty \frac{a_n}{n!}t^n$, we can just insert that into the integral: $$ \int_0^\infty f(t)e^{-t}dt = \int_0^\infty \left(\sum_{n = 0}^\infty \frac{a_n}{n!}t^n\right)e^{-t}dt\\ = \sum_{n = 0}^\infty \int_0^\infty \frac{a_n}{n!}t^ne^{-t}dt\\ = \sum_{n = 0}^\infty \frac{a_n}{n!} \int_0^\infty t^ne^{-t}dt\\ = \sum_{n = 0}^\infty \frac{a_n}{n!} \cdot n!\\ = \sum_{n=0}^\infty a_n = s $$ where the monotone convergence theorem is used to put the sum outside the integral sign, as per Jason Knapp's answer.