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The Von Mangoldt function assigns a natural logarithm weight to each positive power of a prime number. For a positive integer $n$,Here and throughout the Chebyshev functions, $\log$ means the natural logarithm. In particular, $\Lambda (1)=0$: the exponent must be positive, even though the convention in prime power allows exponent zero. The support of a function of $\Lambda$ therefore consists precisely of the prime powers greater than one. ^[ref]

For example, $\Lambda (2)=\Lambda (4)=\Lambda (8)=\log 2$ and $\Lambda (3)=\Lambda (9)=\log 3$, whereas $\Lambda (6)=\Lambda (12)=0$. A composite number can have a nonzero weight: $8$ has just one distinct prime factor. But $12=2^2\cdot 3$ has two distinct prime factors, so its weight is zero. At $p^k$ the weight is $\log p$, not $\log (p^k)=k\log p$. This choice lets successive powers record successive copies of the same prime factor in the Von Mangoldt divisor identity.

The Von Mangoldt function is useful because it connects two views of the prime numbers: the prime factorization of an individual integer, through a divisor sum, and the accumulated weights below a cutoff, through the Second Chebyshev function. The latter turns questions about the distribution of prime numbers into questions about a partial sum of an arithmetic function.

In the average of partial sums $s_0,\dots ,s_N$, the term $a_k$ occurs exactly $N-k+1$ times. Reordering this finite sum gives the displayed identity. Early terms have weights closer to one, and the newest term has weight $1/(N+1)$. The triangular weight expresses the cumulative averaging rule; it is not a claim that one may freely rearrange an infinite series.

For a real exponent $s>1$ and a positive integer $N$, the series truncation error in the defining Riemann zeta function series isThe integral comparison for the positive decreasing function $x^{-s}$ gives the explicit boundsOn each unit interval, the area under $x^{-s}$ lies between its endpoint heights. Summing from the first omitted term gives the lower integral from $N+1$ to infinity and the upper integral from $N$ to infinity. This uses positivity and monotonicity, not an assumption that the last computed term equals the remaining error.

At $N=10$, the bounds are $1/11\le R_{10}(2)\le 1/10$, whereas $1/(3\cdot 11^3)\le R_{10}(4)\le 1/(3\cdot 10^3)$. The errors above fit both intervals. Merely looking at the eleventh term would give $1/121\approx 0.008264$ for squares, far smaller than the actual remaining $0.095166$.

For a chosen absolute error target $\epsilon >0$, the sufficient condition is $N\ge ((s-1)\epsilon {)}^{-1/(s-1)}$, rounded upward to an integer. Thus $\epsilon =10^{-4}$ requires $N=10000$ by this bound for squares but only $N=15$ for fourth powers. This is a guarantee from an upper bound, not a claim that those are the smallest successful term counts.