How do you show that the series 1+sqrt2+root3(3)+...+rootn(n)+... diverges?

1 Answer
Feb 9, 2017

The series is divergent:

sum_(n=1)^oo root(n)n = oo

Explanation:

We can also proceed with direct comparison: as the general term of the series is positive we can take its logarithm:

ln a_n = ln root(n)n = 1/n ln n >= 0 => a_n >= 1

Consider the nth partial sum:

s_n = sum_(i=1)^(i=n) a_i >= sum_(i=1)^(i=n) 1 = n

So for every n:

s_n >= n => lim_(n->oo) s_n = oo