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?

Let us denote the #n^(th)# term of the sequence by:

# u_n = 1/n^3-1/(n+1)^3 #

Then we can denote the partial finite sum by #S_n# so that:

# 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# where:

# S = sum_(r=1)^oo \ u_r #

First we can derive an expression for #S_n#, as we can write the sum of the first #n# terms as follows:

# 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#, we have:

# S = lim_(n rarr oo) S_n #
# \ \ = lim_(n rarr oo) 1 - 1/(n+1)^3 #
# \ \ = 1-0 #
# \ \ = 1 #