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For an arithmetic function $a$, changing the order of the finite summation gives ${\sum }_{d=1}^{n}a(d)\lfloor n/d\rfloor ={\sum }_{t=1}^n{\sum }_{d\le \lfloor n/t\rfloor }a(d)$. Both sides count pairs of positive integers $(d,t)$ with $dt\le n$, assigning weight $a(d)$ to each pair. This expresses a weighted floor sum in terms of a summatory function.

The Mertens function is the summatory function of the Möbius function:For example, $M(1)=1$, $M(2)=0$, $M(3)=M(4)=-1$ and $M(5)=-2$. On nonnegative real inputs it is constant between consecutive integers, with $M(0)=0$. Its jump at a positive integer $n$ is $\mu (n)$, so there is no jump at a nonsquarefree integer.

The Mertens function measures cancellation between square-free integers with an even number and an odd number of distinct prime factors. The estimate $M(x)=o(x)$ says that their signed excess is negligible compared with the total length of the counting interval; through prime number asymptotic from Mertens cancellation, this cancellation implies the Prime number theorem.

Perron's formula recovers coefficient partial sums from a Dirichlet series by the Mellin inversion formula. Suppose $F(s)={\sum }_{n\ge 1}{a}_n{n}^{-s}$ is absolutely convergent at a fixed real $c>0$. For $x>0$,The prime on the sum means that the term with $n=x$ receives weight $1/2$ if $x$ is an integer; for noninteger $x$ it is the ordinary sum over $n<x$. The integral is taken as a symmetric vertical limit, not asserted absolutely convergent on the infinite line.

The scalar truncated Perron kernel estimate tends to $1$ when $x/n>1$, to $0$ when $x/n<1$, and to $1/2$ when $x/n=1$. Absolute convergence on the line allows the coefficient sum to be interchanged with the finite integral and justifies passing to this limit. In analytic number theory, one then shifts the contour through poles and past controlled singularities: their residues supply main terms and error terms in the summatory function. The truncated Perron formula makes the finite-height error explicit.