Find the exact value of the limit
$$\lim_{n\to\infty}\int_0^\infty\left(\sum_{k=0}^n\frac{(-1)^kx^{2k}}{(2k)!}\right)e^{-2x}\;dx.$$
The limit is easily computed by first taking the integral using integration by parts several times, but this question comes from a past real analysis qualifying exam so I'm sure the intended purpose of the problem was to show an understanding of the dominated convergence theorem. Let
$$f_n(x)=\left(\sum_{k=0}^n\frac{(-1)^kx^{2k}}{(2k)!}\right)e^{-2x}$$
then clearly
$$f_n\to\cos(x)e^{-2x},$$ but this does not imply that
$$|f_n|\le\cos(x)e^{-2x}$$ so my question is how can I bound $|f_n|$ in such a way that the dominated convergence theorem can be applied?
Best Answer
Note that for $x \geqslant 0$,
$$\left|\sum_{k=0}^n\frac{(-1)^kx^{2k}}{(2k)!}\right| \leqslant \sum_{k=0}^n\frac{x^{2k}}{(2k)!} < \sum_{k=0}^{2n}\frac{x^{k}}{k!} < e^x$$