Let $f$ be the sinc function, defined by \begin{equation} {\rm{f}}(t)= \begin{cases} \frac{ \sin(2\pi t)}{2\pi t} \qquad &t \not= 0\\ \\ 1\qquad & t =0 \end{cases} \end{equation}
For each integer $n \geq 0$, define \begin{equation}{\rm{f_n}}(t)= \begin{cases} f(t) \qquad &t \in [n,n+1)\\ \\ 0\qquad & otherwise \end{cases} \end{equation} Prove that the series $\sum_{n=0}^\infty f_n$ converges uniformly to $f$.
So for uniform convergence I need to prove that $$\lim_{n \to \infty} \bigg(sup|f(t) - f_n(t)| \bigg ) = 0 $$ So far I found that $$\lim_{t \to \pm \infty}f(t) = 0 $$ and $$\lim_{n \to \infty}f_n(t) = 0 $$ But in general $sup|f(t) - f_n(t)| > 0$ since $f(t) = 1$ for $t = 0$, so Im struggling to see how $f_n$ converges uniformly to $f(t)$.
For each $n\in\Bbb Z_+$ and each $x\in[0,\infty)$,$$f(x)-\sum_{k=0}^nf_k(x)=\begin{cases}0&\text{ if }x<n+1\\\dfrac{\sin(2\pi x)}{2\pi x}&\text{ otherwise.}\end{cases}$$and therefore$$\left|f(x)-\sum_{k=0}^nf_k(x)\right|\leqslant\frac1{2\pi x}\leqslant\frac1{2\pi(n+1)}$$if $x\geqslant n+1$ (and $\left|f(x)-\sum_{k=0}^nf_n(x)\right|=0$ otherwise), So$$\sup_{x\in[0,\infty)}\left|f(x)-\sum_{k=0}^nf_k(x)\right|\leqslant\frac1{2\pi(n+1)}.$$Can you take it from here?