Uniform convergence of $\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ on $\mathbb{R}$? For every $r>0,$ the series $f(x)=\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ converges uniformly on $[-r,r]$. May I know how to prove/disprove that $\sum^{\infty}_{n=0}(-1)^n \frac{x^{2n}}{n!}$ converges uniformly on $\mathbb{R}$ ?
Thank you. 

Generally, if $(p_n)$ is a sequence of polynomials converging uniformly on $\mathbf{R}$ to a function $f$, then $f$ itself is a polynomial. In particular, a power series converges uniformly on $\mathbf{R}$ if and only if the series has only finitely many non-zero terms.
Briefly, $(p_n)$ is uniformly Cauchy on $\mathbf{R}$, so there exists an $N$ such that if $n \geq N$, then $|p_n(x) − p_N(x)| < 1$ for all real $x$. That is, $p_n  − p_N$ is a bounded polynomial, i.e., a constant.
