In my homework I am trying to calculate $$\sum_{n=0}^\infty \frac{1}{n!} \int_{[0,1]} x^n \frac{1}{e^{2x}} \, d\lambda(x)$$
Using the constant $\frac{1}{n!}$ I get: $$=\sum_{n=0}^\infty \int_{[0,1]} \frac{1}{n!}\cdot x^n \frac{1}{e^{2x}}$$ and switching sum and integral: $$= \int_{[0,1]} \sum_{n=0}^\infty \frac{x^n}{n!} \frac{1}{e^{2x}}$$ and using that $\sum_{n=0}^\infty x^n/n! = e^x$ I get: $$= \int_{[0,1]} e^x \frac{1}{e^{2x}}= \int_{[0,1]} \frac{e^x}{e^{2x}}= \int_{[0,1]} \frac{1}{e^{2x-x}}$$ $$= \int_{[0,1]} \frac{1}{e^{x}}= \int_{[0,1]} \frac{1}{e^{x}}=-\frac{1}{e^x}\bigg|_0^1=-\frac{1}{e^1}-(-\frac{1}{e^0})$$ and final result: $$=-\frac{1}{e}+1$$
However this only works because I can switch the integral and sum and I don't know how to argue for this and therefore I have some doubts on the final result
Any help/hint would be appreciated
Your answer is correct and the method can be justified in many ways. One way is to note that $ \sum\limits_{n=0}^{m} \frac 1 {n!} x^{n} \frac 1 {e^{2x}}$ is a sequence of non-negative measurable functions increasing to $ \sum\limits_{n=0}^{\infty} \frac 1 {n!} x^{n} \frac 1 {e^{2x}}$ at every point as $m$ increases to $\infty$. An application of Monotone Convregenec Theorem jsutifies interchange of sum and intergal in your argument.