Today when I was studying Apostol's Analytical Number theory, I came to know about the formula $\sum\limits_{d|n}\mu(d)=1$ if $n=1$ and $0$ otherwise. I understood the technique and then using the same approach I was trying to find out if it is possible to derive any closed formula for $$\sum\limits_{d\nmid n}\mu(d)$$
Can someone help me how to evaluate it ?
You won't find a closed formula. That same sum, however, is interesting in terms of its asymptotic behaviour as $n \to \infty $. The Moebius function is zero at at integers divisible by the square of a prime, as you surely know, and at squarefree integers it is either $1$ or $-1$ depending on whether it has an even or odd amount of prime factors. Hence, if there are roughly as many numbers with an even amount of primes factors as there are with an odd amount, wed expect that the $1$s and $-1$s would cancel on average and we would have $$ \sum_{k \leq n} \mu (k) = o(n) $$ This is indeed the case. Interestingly, this statement is equivalent to the Prime Number Theorem. Indeed, in terms of the Riemann Zeta Function, the statement is equivalent to proving that there are no zeroes on the line $Re(s)=1$. If you assume the Riemann Hypothesis, you get the far better estimate $$ \sum_{k \leq n} \mu(k) = O(x^{1/2 + \epsilon}) $$ for any $\epsilon > 0$