I'am preparing for mathematical analysis exam. I have one excercise to do; calculate: $$ \int^\infty_3 \sum^\infty_{n=0} \frac{(-2x)^{n+1}}{n!} \,dx $$
I don't know even wehere to start... I suppose here's some 'popular' sum etc. but how to approach this without knowing it?
You should recognize the resemblance between your series $\displaystyle\sum^\infty_{n=0} \frac{(-2x)^{n+1}}{n!}$ and the well known series $\displaystyle \sum_{n=0}^\infty \frac{a^n}{n!} = e^a.$
You have $a^{n+1}$ in the numerator instead of $a^n,$ so write $$ \sum_{n=0}^\infty \frac{a^{n+1}}{n!} = \sum_{n=0}^\infty a\cdot \frac{a^n}{n!}. $$ The factor $a$ does not change as $n$ goes from $0$ to $\infty,$ so it can be pulled out: $$ = a\sum_{n=0}^\infty \frac{a^n}{n!} = ae^a. $$ Then you have $$ \int_3^\infty (-2x)e^{-2x} \, dx $$ Then you can integrate by parts.