@srestha,@arjun sir

Please check two points:-

$Point1$

If i consider merge sort recreance

$T(n) = 2T(n/2) +0(n) +c $

Each call to merge will solve two subproblems,and then we have $(o(n)$ time to merge them by a fxn call.Now it will also do some constant operations which are represented by $c$.

But we take $n+c=n$ and write it as :-

$T(n) = 2T(n/2) +0(n) $

so, why cant i remove c here as @yg92 has aksed?

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$Point2$

If i cannot be done then i think we can rewrite the given recreance as:-

$T(n)=T(n−1)+1/n+c$

$T(n)=T(n−1)+c1+c $// As 1/n is always less than 1,means some constant(c1)

$T(n)=T(n−1)+k; $// k is a constant ,k=c1+c

Now solve it we get $o(n)$ times.Please check once,?

But we cannot remove c,as i did in point 1,because $1/n$ can be less than 1(it cannot be even equal because if 1/n=1 means n=1 means base condition),but constant will always be +ve,may be one or more,so constant will dominate.

More strongly i can say that $1/n$ is strictly less than constant