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

Feb 9, 2017

The series is divergent:

${\sum}_{n = 1}^{\infty} \sqrt[n]{n} = \infty$

Explanation:

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

$\ln {a}_{n} = \ln \sqrt[n]{n} = \frac{1}{n} \ln n \ge 0 \implies {a}_{n} \ge 1$

Consider the $n$th partial sum:

${s}_{n} = {\sum}_{i = 1}^{i = n} {a}_{i} \ge {\sum}_{i = 1}^{i = n} 1 = n$

So for every $n$:

${s}_{n} \ge n \implies {\lim}_{n \to \infty} {s}_{n} = \infty$