Mathematical Methods for Physics and Engineering - Matematica.NET

4.2 SUMMATION OF SERIES4.2.5 Series involving natural numbersSeries consisting of the natural numbers 1, 2, 3, ..., or the square or cube of thesenumbers, occur frequently and deserve a special mention. Let us first considerthe sum of the first N natural numbers,N∑S N =1+2+3+···+ N = n.This is clearly an arithmetic series with first term a = 1 and common differenced = 1. Therefore, from (4.2), S N = 1 2N(N +1).Next, we consider the sum of the squares of the first N natural numbers:N∑S N =1 2 +2 2 +3 2 + ...+ N 2 = n 2 ,which may be evaluated using the difference method. The nth term in the seriesis u n = n 2 , which we need to express in the form f(n) − f(n − 1) for some functionf(n). Consider the functionf(n) =n(n + 1)(2n +1) ⇒ f(n − 1) = (n − 1)n(2n − 1).For this function f(n) − f(n − 1) = 6n 2 , and so we can writeu n = 1 6[f(n) − f(n − 1)].Therefore, by the difference method,S N = 1 6 [f(N) − f(0)] = 1 6N(N + 1)(2N +1).Finally, we calculate the sum of the cubes of the first N natural numbers,N∑S N =1 3 +2 3 +3 3 + ···+ N 3 = n 3 ,again using the difference method. Consider the functionf(n) =[n(n +1)] 2 ⇒ f(n − 1) = [(n − 1)n] 2 ,for which f(n) − f(n − 1) = 4n 3 . Therefore we can write the general nth term ofthe series asu n = 1 4[f(n) − f(n − 1)],and using the difference method we findS N = 1 4 [f(N) − f(0)] = 1 4 N2 (N +1) 2 .Note that this is the square of the sum of the natural numbers, i.e.(N∑N 2∑n 3 = n).n=1121n=1n=1n=1n=1