Proving $\lim_{n\to\infty}\sum_{k=n}^{\infty}\Big(\frac{1}{n\left\lfloor\frac kn\right\rfloor}-\frac1k\Big)=\gamma$

Question

I remember arriving at the following equality: $$\lim_{n\to\infty}\sum_{k=n}^{\infty}\left(\frac{1}{n\left\lfloor\frac kn\right\rfloor}-\frac1k\right)=\gamma$$ where $\gamma$ denotes the Euler-Mascheroni constant. However, I cannot find back where I wrote its proof, and I am now having a hard time reconstructing it. I don't remember any clues as to how I derived this. I am now not even able to prove it converges (which it does, but really slowly). How can I prove this? Thanks in advance.

Answer 1

Fix $n\geq 1$, and consider an arbitrary integer $L \geq 1$. Consider $$\begin{align} A_n(L) &= \sum_{k=n}^{Ln-1} \left( \frac{1}{n\left\lfloor \frac{k}{n}\right\rfloor }-\frac{1}{k} \right) = \sum_{\ell=1}^{L-1} \sum_{k=\ell n}^{(\ell+1)n-1} \left( \frac{1}{n\left\lfloor \frac{k}{n}\right\rfloor}-\frac{1}{k} \right) \\ &= \sum_{\ell=1}^{L-1} \sum_{k=\ell n}^{(\ell+1)n-1} \left( \frac{1}{\ell n }-\frac{1}{k} \right) = \sum_{\ell=1}^{L-1} \frac{1}{\ell}-\sum_{\ell=1}^{L-1} \sum_{k=\ell n}^{(\ell+1)n-1} \frac{1}{k} \\ &= H_{L-1}-\sum_{k=n}^{Ln-1} \frac{1}{k} = H_{L-1} - (H_{Ln-1} - H_{n-1}) \\ &= \ln L + \gamma + o_L(1) - (\ln L + \ln n + \gamma + o_L(1) - (\ln n + \gamma + o_n(1))) \\ &= \gamma + o_n(1) + o_L(1) \end{align}$$ where I use the notation $o_N(1)$ for terms that go to $0$ when $N \to \infty$, and used the asymptotic expansion of the harmonic series: $$H_N = \ln N + \gamma + o_N(1)$$ (as well as the fact that $\ln(N-1) = \ln N + o_N(1)$) . It follows that the sum of non-negative terms $$ \sum_{k=n}^{\infty} \left( \frac{1}{n\left\lfloor \frac{k}{n}\right\rfloor }-\frac{1}{k} \right) $$ exists and is equal to $$ \sum_{k=n}^{\infty} \left( \frac{1}{n\left\lfloor \frac{k}{n}\right\rfloor }-\frac{1}{k} \right) = \lim_{L\to\infty} A_n(L) = \gamma + o_n(1). $$ This in turn shows that $$ \sum_{k=n}^{\infty} \left( \frac{1}{n\left\lfloor \frac{k}{n}\right\rfloor }-\frac{1}{k} \right) \xrightarrow[n\to\infty]{} \gamma. $$

Answer 2

$n\lfloor\dfrac kn\rfloor$ makes $n$ copies of the naturals times $n$ in a row.

So the partial sums from $k=n$ to $nm+n-1$ converge to $n\sum_{k=1}^m\dfrac1{nk}-\sum_{k=n}^{nm+n-1}\dfrac1k=$ $n\dfrac{\ln(m)+\gamma}n-(\ln(nm+n-1)-\ln(n))=\ln(m)-\ln\left(m+1-\frac1n\right)+\gamma$.