In the Nash bargaining problem with a finite horizon, the number of periods is changed from $2$ to $2$n$.$ Two consecutive periods are referred to as a super$-$period. For instance, period $1$ and period $2$ are called super$-$period $1,$ periods $3$ and $4$ are called super$-$period $2,$ and so on$.$ In general, super$-$period i$,$ where i ranges from $1$ to n$,$ contains two periods. The first period is period $2$i$-1$ and the second period is period $2$i$.$

Node A proposes a split in the first period of super$-$periods and Node B proposes a split in the second period of super$-$periods. The split proposed by node A in super$-$period i is denoted by

$,$ where i ranges from $1$ to n$.$ Similarly, the split proposed by node B in super$-$period i is denoted by a$)$ Derive $($a$1($i$),$ b$1($i$))$ and $($a$2($i$),$ b$2($i$))$ for i $=1,2,...,$ n$.$

b$)$ Show that as n goes to infinity, the limits of the splits derived in part $($a$)$ are