Suppose that $f_0 = 1$ and for $n \geq 1$, $$f_n(x) = \sum_{k=0}^{n-1} \frac{x^{2k+1}}{\prod\limits_{i=0}^k(2i+1)} + \frac{x^{2n}}{\prod\limits_{i=1}^n (2i)}.$$
I'm wondering whether this sequence $(f_n)$ converges uniformly on $[-b,b]$ where $b > 0$. I'm trying to show that it's Cauchy in $C[-n,n]$, but this seems to be really difficult. Is there any hint or suggestion?
Yes, it does. All that you need to prove is that the series$$\sum_{k=0}^\infty\frac{x^{2k+1}}{\prod_{i=0}^k(2i+1)}$$and the sequence$$\left(\frac{x^{2n}}{\prod_{i=0}^n(2i)}\right)_{n\in\Bbb N}$$both converge uniformly on $[-b,b]$. The first assertion is true by the Weierstrass $M$-test: $\sup_{x\in[-b,b]}\left|\frac{x^{2k+1}}{\prod_{i=0}^k(2i+1)}\right|=\frac{b^{2k+1}}{\prod_{i=0}^k(2i+1)}$ and the series $\sum_{k=0}^\infty\frac{b^{2k+1}}{\prod_{i=0}^k(2i+1)}$ converges (by, say, the ratio test). And, on the other hand, $\sup_{x\in[-b,b]}\left|\frac{x^{2n}}{\prod_{i=0}^n(2i)}\right|\leqslant\frac{b^{2n}}{\prod_{i=0}^n(2i)}$ and $\lim_{n\to\infty}\frac{b^{2n}}{\prod_{i=0}^n(2i)}=0$.