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

Question

1816 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-09T16:53:46

The series is divergent:

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

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 #n#th 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#