How do you determine if #a_n=(1-1/8)+(1/8-1/27)+(1/27-1/64)+...+(1/n^3-1/(n+1)^3)+...# converge and find the sums when they exist?
1 Answer
The sum of the first
# S_n = 1 - 1/(n+1)^3 #
The sum to infinity is given by:
# S = 1 #
Explanation:
Let us denote the
# u_n = 1/n^3-1/(n+1)^3 #
Then we can denote the partial finite sum by
# S_n = sum_(r=1)^n \ u_r #
# \ \ \ \ = (1-1/8)+(1/8-1/27)+... + (1/n^3-1/(n+1)^3)#
And then the infinite sum, by
# S = sum_(r=1)^oo \ u_r #
First we can derive an expression for
# S_n = u_1 + u_2 + ... u_n #
# \ \ \ \ \ = {1-1/8} + #
# \ \ \ \ \ \ \ \ \ \ {1/8-1/27} + #
# \ \ \ \ \ \ \ \ \ \ {1/27-1/64} + #
# \ \ \ \ \ \ \ \ \ \ vdots #
# \ \ \ \ \ \ \ \ \ \ {1/n^3-1/(n+1)^3 } #
This is a difference sum, and we can see that almost all terms will cancel with other terms:
# S_n = {1-cancelcolor(red)(1/8)} + #
# \ \ \ \ \ \ \ \ \ \ {cancelcolor(red)(1/8)-cancelcolor(blue)(1/27)} + #
# \ \ \ \ \ \ \ \ \ \ {cancelcolor(blue)(1/27)-cancel(1/64)} + #
# \ \ \ \ \ \ \ \ \ \ vdots #
# \ \ \ \ \ \ \ \ \ \ {(cancel(1/n^3)-1/(n+1)^3 } #
After which we are left with:
# S_n = 1 - 1/(n+1)^3 #
And for the sum,
# S = lim_(n rarr oo) S_n #
# \ \ = lim_(n rarr oo) 1 - 1/(n+1)^3 #
# \ \ = 1-0 #
# \ \ = 1 #