If $(X_n, d_n)$ is a family of complete metric spaces, prove that $(X,d)$ is complete Let $(X_n, d_n)$ be a family of complete metric spaces. Prove that $(X,d)$ is a complete metric space.
$$
X = \prod_{n=1}^{\infty} X_n = \{(x_n)_{n \in \mathbb{N}} : x_j \in X_j \}
$$
Furthermore let $(\gamma_n)_{n \in \mathbb{N}} \subset (0, \infty)$ be a seq. s.t $\sum_{n=1}^{ \infty} \gamma_n < \infty$. For $x,y \in X$ where $x = (x_n)_{n \in \mathbb{N}}$, $y = (y_n)_{n \in \mathbb{N}}$ define 
$$
d(x,y) = \sum_{n=1}^{\infty} \gamma_n \frac{d_n(x_n,y_n)}{1+d_n(x_n,y_n)}
$$
Where $d_n$ is the metric on $X_n$
This is what I have come up with.
Let $(x^m)_{m \in \mathbb{N}}$ be Cauchy seq in $X$. Thus we know that $(x_n^m)_{m \in \mathbb{N}}$ is Cachy in $X_n$ (I proved that earlies problem). Hence $(x_n^m)_{m \in \mathbb{N}}$ converges to some $x_n \in X_n$. Then
$$
d(x_n^m, x_n) = \sum_{n=1}^{\infty} \gamma_n \frac{d_n(x_n^m, x_n)}{1 + d_n(x_n^m,x_n)} < \sum_{n=1}^{\infty} \gamma_n \frac{\epsilon}{1 + \epsilon} = \frac{\epsilon}{1 + \epsilon} \sum_{n=1}^{\infty} \gamma_n < \frac{\epsilon}{1 + \epsilon} \cdot M < \epsilon \cdot M
$$
Where $M = \sum_{n=1}^{\infty} \gamma_n$ since $\sum_{n=1}^{\infty} \gamma_n < \infty$
 A: The step $$\tag{*}  \sum_{n=1}^{\infty} \gamma_n \frac{d_n\left(x_n^m, x_n\right)}{1 + d_n\left(x_n^m, x_n\right)} < \sum_{n=1}^{\infty} \gamma_n \frac{\epsilon}{1 + \epsilon}$$ 
need to be more detailed; in particular, we have to precise for which $m$'s it holds. It seems that you use the fact that $d_n\left(x_n^m, x_n\right) \lt\varepsilon$ for $m$ large enough, but the "large enough" may depend on the considered $m$. Nevertheless, an inequality similar to (*) does hold for $m$ large enough. Indeed, fix a positive $\epsilon$. There exists $M$ such that if $m,m'\geqslant M$, then $d\left(x^m,x^{m'}  \right) \lt\epsilon$. We thus have for any fixed $N$ and $m,m'\geqslant M$ that 
$$\sum_{n=1}^{N} \gamma_n \frac{d_n\left(x_n^m, x_n^{m'}  \right)}{1 + d_n\left(x_n^m, x_n\right)} \lt \epsilon.$$
In particular, letting $m'$ going to infinity, we get for any $N\geqslant 1$
$$\sum_{n=1}^{N} \gamma_n \frac{d_n\left(x_n^m, x_n\right)}{1 + d_n\left(x_n^m, x_n\right)} \leqslant \epsilon .$$
 This implies by letting $N$ going to $+\infty$ that $d\left(x^{(m)},x\right)\leqslant \varepsilon$ for each $m\geqslant M$.  
