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Sum of Natural Numbers (second proof and extra footage) includes demonstration of Euler's method. What do we get if we sum all the natural numbers? response to comments about video by Tony Padilla; Related article from New York Times; Why –1/12 is a gold nugget follow-up Numberphile video with Edward Frenkel
For example, the sum of the first n natural numbers can be denoted as ∑ i = 1 n i {\displaystyle \sum _{i=1}^{n}i} For long summations, and summations of variable length (defined with ellipses or Σ notation), it is a common problem to find closed-form expressions for the result.
The n-th harmonic number, which is the sum of the reciprocals of the first n positive integers, is never an integer except for the case n = 1. Moreover, József Kürschák proved in 1918 that the sum of the reciprocals of consecutive natural numbers (whether starting from 1 or not) is never an integer.
A natural number is a Dudeney root if it is a fixed point for ,, which occurs if , =. The natural number m = n p {\displaystyle m=n^{p}} is a generalised Dudeney number , [ 1 ] and for p = 3 {\displaystyle p=3} , the numbers are known as Dudeney numbers .
Since a = n(n + 1)/2, these formulae show that for an odd power (greater than 1), the sum is a polynomial in n having factors n 2 and (n + 1) 2, while for an even power the polynomial has factors n, n + 1/2 and n + 1.
The harmonic number with = ⌊ ⌋ (red line) with its asymptotic limit + (blue line) where is the Euler–Mascheroni constant.. In mathematics, the n-th harmonic number is the sum of the reciprocals of the first n natural numbers: [1] = + + + + = =.
The function q(n) gives the number of these strict partitions of the given sum n. For example, q(3) = 2 because the partitions 3 and 1 + 2 are strict, while the third partition 1 + 1 + 1 of 3 has repeated parts. The number q(n) is also equal to the number of partitions of n in which only odd summands are permitted. [20]
The number of distinct prime factors of a pronic number is the sum of the number of distinct prime factors of n and n + 1. If 25 is appended to the decimal representation of any pronic number, the result is a square number, the square of a number ending on 5; for example, 625 = 25 2 and 1225 = 35 2. This is so because