Let $C^{\infty}(\mathbb{R})$ be the space of all infinitely differentiable complex-valued functions on $\mathbb{R}$. Define the metric \begin{align} d(f,g) = \sum_{n,m=0}^{\infty} 2^{-n-m}\frac{\sup_{x\in\mathbb{R}:|x|\leq n} |f^{(m)}(x)-g^{(m)}(x)|}{1+\sup_{x\in\mathbb{R}:|x|\leq n} |f^{(m)}(x)-g^{(m)}(x)|}. \end{align} Show that $(C^{\infty}(\mathbb{R}),d)$ is complete.
My attempt: Let $(f_k)_{k\in\mathbb{N}}$ be a Cauchy sequence in $(C^{\infty}(\mathbb{R}),d)$. Then we know that $(f_k^{(m)}(x))_{k\in\mathbb{N}}$ is a Cauchy sequence in $\mathbb{C}$ for each $m\in\mathbb{N}_{0}$ and each $x\in\mathbb{R}$. Since $\mathbb{C}$ is complete $(f_k^{(m)}(x))_{k\in\mathbb{N}}$ convergence to some $g_{m}(x)\in\mathbb{C}$ for each $m\in\mathbb{N}_{0}$ and each $x\in\mathbb{R}$.
Next, I want to show that $(f_k^{(m)})_{k\in\mathbb{N}}$ convergence uniformly to $g_{m}$ for each $m\in\mathbb{N}_{0}$ but not sure how to do it. At least I want to show that $g_{m}$ is continuous.
Since this seems to be where OP is struggling, I’m writing answer to prove the following: if on each $[-n, n]$, $f_k^{(m)} \to g^{(m)}$ uniformly, then $d(f_k, g) \to 0$.
Let $\epsilon > 0$. Then there exists $N > 0$ s.t. $\sum_{n \geq N \vee m \geq N} 2^{-n-m} < \epsilon/2$ (here, $\vee$ means or). Thus,
$$d(f_k, g) < \sum_{n, m = 0}^{N-1} 2^{-n-m}\frac{\sup_{x\in\mathbb{R}:|x|\leq n} |f^{(m)}(x)-g^{(m)}(x)|}{1+\sup_{x\in\mathbb{R}:|x|\leq n} |f^{(m)}(x)-g^{(m)}(x)|} + \frac{\epsilon}{2}$$
Since for each $n$ and $m$, on $[-n, n]$, $f_k^{(m)} \to g^{(m)}$ uniformly, there exists $K > 0$ s.t., whenever $k \geq K$,
$$\frac{\sup_{x\in\mathbb{R}:|x|\leq n} |f^{(m)}(x)-g^{(m)}(x)|}{1+\sup_{x\in\mathbb{R}:|x|\leq n} |f^{(m)}(x)-g^{(m)}(x)|} < \frac{2^{n + m}}{2N^2}\epsilon$$
For all $n, m \leq N$. Thus, whenever $k \geq K$,
$$\begin{split} d(f_k, g) &< \sum_{n, m = 0}^{N-1} 2^{-n-m}\frac{\sup_{x\in\mathbb{R}:|x|\leq n} |f^{(m)}(x)-g^{(m)}(x)|}{1+\sup_{x\in\mathbb{R}:|x|\leq n} |f^{(m)}(x)-g^{(m)}(x)|} + \frac{\epsilon}{2}\\ &< \sum_{n, m = 0}^{N-1} 2^{-n-m} \frac{2^{n + m}}{2N^2}\epsilon + \frac{\epsilon}{2}\\ &= \epsilon \end{split}$$
So $d(f_k,g) \to 0$, as required.