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

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

Linear isometry

Limit with harmonic number

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