Mathematical Methods for Physics and Engineering : A Comprehensive Guide

4.2 SUMMATION OF SERIES

4.2.5 Series involving natural numbers

Series consisting of the natural numbers 1, 2, 3,..., or the square or cube of these

numbers, occur frequently and deserve a special mention. Let us first consider

the sum of the firstNnatural numbers,

SN=1+2+3+···+N=

∑N

n=1

n.

This is clearly an arithmetic series with first terma= 1 and common difference

d= 1. Therefore, from (4.2),SN=^12 N(N+1).

Next, we consider the sum of the squares of the firstNnatural numbers:

SN=1^2 +2^2 +3^2 +...+N^2 =

∑N

n=1

n^2 ,

which may be evaluated using the difference method. Thenth term in the series

isun=n^2 , which we need to express in the formf(n)−f(n−1) for some function

f(n). Consider the function

f(n)=n(n+ 1)(2n+1) ⇒ f(n−1) = (n−1)n(2n−1).

For this functionf(n)−f(n−1) = 6n^2 , and so we can write

un=^16 [f(n)−f(n−1)].

Therefore, by the difference method,

SN=^16 [f(N)−f(0)] =^16 N(N+ 1)(2N+1).

Finally, we calculate the sum of the cubes of the firstNnatural numbers,

SN=1^3 +2^3 +3^3 +···+N^3 =

∑N

n=1

n^3 ,

again using the difference method. Consider the function

f(n)=[n(n+1)]^2 ⇒ f(n−1) = [(n−1)n]^2 ,

for whichf(n)−f(n−1) = 4n^3. Therefore we can write the generalnth term of

the series as

un=^14 [f(n)−f(n−1)],

and using the difference method we find

SN=^14 [f(N)−f(0)] =^14 N^2 (N+1)^2.

Note that this is the square of the sum of the natural numbers, i.e.

∑N

n=1

n^3 =

(N

∑

n=1

n

) 2

.