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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
In mathematics, summation is the addition of a sequence of numbers, called addends or summands; the result is their sum or total.Beside numbers, other types of values can be summed as well: functions, vectors, matrices, polynomials and, in general, elements of any type of mathematical objects on which an operation denoted "+" is defined.
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.
There are two popular ways to define the sum of two natural numbers a and b. If one defines natural numbers to be the cardinalities of finite sets, (the cardinality of a set is the number of elements in the set), then it is appropriate to define their sum as follows: Let N(S) be the cardinality of a set S.
The algorithm performs summation with two accumulators: sum holds the sum, and c accumulates the parts not assimilated into sum, to nudge the low-order part of sum the next time around. Thus the summation proceeds with "guard digits" in c , which is better than not having any, but is not as good as performing the calculations with double the ...
In number theory, Ramanujan's sum, usually denoted c q (n), is a function of two positive integer variables q and n defined by the formula c q ( n ) = ∑ 1 ≤ a ≤ q ( a , q ) = 1 e 2 π i a q n , {\displaystyle c_{q}(n)=\sum _{1\leq a\leq q \atop (a,q)=1}e^{2\pi i{\tfrac {a}{q}}n},}
It is used to prove Kronecker's lemma, which in turn, is used to prove a version of the strong law of large numbers under variance constraints. It may be used to prove Nicomachus's theorem that the sum of the first n {\displaystyle n} cubes equals the square of the sum of the first n {\displaystyle n} positive integers.
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.