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Harmonic limit

Let \mathcal{H}_n denote the n – th harmonic number. Evaluate the limit

    \[\ell =\lim_{n \rightarrow +\infty} \left ( \frac{1}{\ln n} \sum_{k=1}^{n} \frac{\mathcal{H}_k}{k} - \frac{\ln n}{2} \right )\]

Solution

Lemma: It holds that \displaystyle \sum_{k=1}^{n} \frac{\mathcal{H}_k}{k} = \frac{\mathcal{H}_n^2+\mathcal{H}_n^{(2)}}{2}.

Proof: We have successively:

    \begin{align*} 2\sum_{k=1}^n \frac{\mathcal{H}_k}{k} &= 2\sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} \\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{j=1}^n \sum_{k=j}^n \frac{1}{jk}\\ &= \sum_{k=1}^n \sum_{j=1}^k \frac{1}{jk} + \sum_{k=1}^n \sum_{j=k}^n \frac{1}{jk}\\ &= \sum_{k=1}^n \sum_{j=1}^n \frac{1}{jk} + \sum_{k=1}^n \frac{1}{k^2} \\ &= \left(\sum_{k=1}^n \frac{1}{k} \right)^2 + \sum_{k=1}^n \frac{1}{k^2} \\ &= \mathcal{H}_n^2+ \mathcal{H}^{(2)}_n \\ \end{align*}

Thus,

    \begin{align*} \ell &= \lim_{n \rightarrow +\infty} \left ( \frac{1}{\ln n} \sum_{k=1}^{n} \frac{\mathcal{H}_k}{k} - \frac{\ln n}{2} \right )\\ &=\lim_{n \rightarrow +\infty} \left ( \frac{\mathcal{H}_n^2 +\mathcal{H}_n^{(2)}}{2\ln n} - \frac{\ln n}{2} \right ) \\ &=\lim_{n \rightarrow +\infty} \left ( \frac{\left ( \gamma + \ln n + \mathcal{O} \left ( \frac{1}{n} \right ) \right )^2 + \left ( \zeta(2) - \frac{1}{n} + \mathcal{O}\left ( \frac{1}{n^2} \right ) \right )}{2\ln n} - \frac{\ln n}{2} \right ) \\ &=\gamma \end{align*}

 

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