\[\zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s}\]
\[\sum_{i=2}^{3} > \]
Let n be the no of array elements \newline
Let \ c(n) be the no of comparisions \newline
\newline
When $n=1$ $c(n)$ is also 1
\newline
The time taken to solve array of size $n/2$ would be $T(n/2)$
\newline
A comparison is made only when i=j, hence there would be two comparisons \newline
one for each for the divided arrays
\newline
There would be no operations for combining the result
\newline

Hence the recurrence relation would be:\newline

\[T(n) \le \begin{cases}
1 & \text{if n=1}  \\
2T(n/2)+2& \text{if n>1}  
\end{cases}\]
Solving the recurrence
\[n = 2^{k}\]
\[T(1) = 1\]