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Determine the convergence of the series 1 / n(1 + 1/2 + … + 1/n)

by RoRi

Consider the series

    \[ \sum_{n=1}^{\infty} \frac{1}{n \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)}. \]

Determine whether the series converges or diverges. If it converges, determine whether it converges conditionally or absolutely.

The series diverges.

Proof. We use the limit comparison test with the series \sum \frac{1}{n \log n}. Let

    \[ a_n = \frac{1}{n \log n}, \qquad b_n = \frac{1}{n \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)}. \]

Then, taking the limit we have

    \begin{align*} \lim_{n \to \infty} \frac{a_n}{b_n} &= \lim_{n \to \infty} \frac{n \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)}{n \log n} \\[9pt] &= \lim_{n \to \infty} \frac{1 + \frac{1}{2} +\cdots + \frac{1}{n} }{\log n} \\[9pt] &= 1. \end{align*}

(The final equality follows from Example 3 on page 405 of Apostol, which tells us that this limit is 1.) Therefore, \sum a_n and \sum b_n either both converge or both diverge. But since

    \[ \sum_{n=1}^{\infty} \frac{1}{n \log n} \]

diverges we can then conclude that

    \[ \sum_{n=1}^{\infty} \frac{1}{n \left( 1 + \frac{1}{2} + \cdots + \frac{1}{n} \right)} \]

diverges as well. \qquad \blacksquare

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