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Limit with harmonic number

Let \mathcal{H}_n denote the n-th harmonic number. Prove that

    \[\lim_{n \rightarrow +\infty} n \left(\mathcal{H}_n -\log n -\gamma \right) = \frac{1}{2}\]

Solution

Using the well known asymptotic formula for the n – th harmonic number

    \[\mathcal{H}_n \sim \log n + \gamma + \frac{1}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right )\]

we conclude that

    \begin{align*} n \left ( \mathcal{H}_n - \log n - \gamma \right ) & \sim n \left [ \log n + \gamma + \frac{1}{2n} + \mathcal{O} \left ( \frac{1}{n^2} \right ) - \log n - \gamma \right ] \\ &\sim \frac{1}{2} + \mathcal{O} \left ( \frac{1}{n} \right ) \\ &\longrightarrow \frac{1}{2} \end{align*}

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