Is $\hat{G}$ is complete with respect to the induced topology of $G$? For a topological group $G$ and  a given fundamental system of neighbourhoods of $G$ we can define the completion of G and we call it $\hat{G}$. The induced fundamental system of neighbourhoods of $\hat{G}$ is given by this  Topology induced by the completion of a topological group. 

Then Can we say that $\hat{G}$ is complete ? i.e., every Cauchy sequence in $\hat{G}$ is convergent ?

For this we assume that $\{z_n\}$ be any Cauchy sequence in $\hat{G},$ then given any open neighbourhood $\tilde{N}$ of $\hat{G}$ there exists an integer $k$ such that whenever $m,n \geq k,$ $z_m-z_n \in \tilde{N}.$ Then how can I show that $\{z_n\}$ is convergent ? That is we are looking for an element $s \in \hat{G}$ such that for any neighbourhood $\hat{P}$ of $\hat{G},$ $z_n \in s+\hat{P}$ for large $n$. In general will it hold ? I need some help. Thanks.
 A: For simplicity, I'll ignore the very accurate point made in the comments that defining completeness in terms of sequences isn't sufficient in general, and assume that the topology on $G$ is first-countable, and therefore $\hat G$ is first-countable as well.
First, we need to correct your definition of Cauchy sequences and of convergence. For example, in $\Bbb R, \left\{\frac 1n\right\}$ is a Cauchy sequence (as is any convergent sequence), and $(1,2)$ is a neighborhood in $\Bbb R$, but there is not a $k \in \Bbb N$ such that for $n, m > k, \left(\frac 1m - \frac 1n\right) \in (1,2)$.
$\tilde N$ needs to be a neighborhood of $0$, not just any neighborhood of $\hat G$. Similarly, $\hat P$ is also a neighborhood of $0$ (and so $s + \hat P$ is a neighborhood of $s$).
Since $\{z_n\}_{n\in\Bbb N} \subset \hat G$, for each $n, z_n$ is some Cauchy sequence $\{z_{nm}\}_{m\in\Bbb N}$ in $G$. You can use the fact that $\{z_n\}_{n\in\Bbb N}$ is Cauchy in $\hat G$ to show that the diagonal sequence $\{z_{nn}\}_{n\in \Bbb N}$ is Cauchy in $G$. 
Then $s = \{z_{nn}\}_{n\in \Bbb N}$.
