Does #a_n = (n+1)^(1/3)-n^(1/3)# converge?

1 Answer
Apr 10, 2017

As #n->oo# the individual terms tend to #0#, but the sum tends to #oo#.

Explanation:

Given:

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

Use the difference of cubes identity:

#a^3-b^3 = (a-b)(a^2+ab+b^2)#

with #a=(n+1)^(1/3)# and #b=n^(1/3)# as follows:

#lim_(n->oo) a_n = lim_(n->oo) ((n+1)^(1/3)-n^(1/3))#

#color(white)(lim_(n->oo) a_n) = lim_(n->oo) (((n+1)^(1/3)-n^(1/3))((n+1)^(2/3)+(n+1)^(1/3)n^(1/3)+n^(2/3)))/((n+1)^(2/3)+(n+1)^(1/3)n^(1/3)+n^(2/3))#

#color(white)(lim_(n->oo) a_n) = lim_(n->oo) ((n+1)-n)/((n+1)^(2/3)+(n+1)^(1/3)n^(1/3)+n^(2/3))#

#color(white)(lim_(n->oo) a_n) = lim_(n->oo) 1/((n+1)^(2/3)+(n+1)^(1/3)n^(1/3)+n^(2/3))#

#color(white)(lim_(n->oo) a_n) = 0#

We also find:

#sum_(n=1)^N a_n = sum_(n=1)^N ((n+1)^(1/3)-n^(1/3))#

#color(white)(sum_(n=1)^N a_n) = sum_(n=1)^N (n+1)^(1/3)-sum_(n=1)^N n^(1/3)#

#color(white)(sum_(n=1)^N a_n) = sum_(n=2)^(N+1) n^(1/3)-sum_(n=1)^N n^(1/3)#

#color(white)(sum_(n=1)^N a_n) = (N+1)^(1/3)+color(red)(cancel(color(black)(sum_(n=2)^N n^(1/3))))-color(red)(cancel(color(black)(sum_(n=2)^N n^(1/3)))) - 1^(1/3)#

#color(white)(sum_(n=1)^N a_n) = (N+1)^(1/3)-1#

Hence:

#sum_(n=1)^oo a_n = lim_(N->oo) ((N+1)^(1/3)-1) = oo#