Calculating Harmonic Numbers

Question

I'm hoping someone can confirm if I did this right, is the expression for calculating the nth harmonic number as written below?

$$ H_n=\gamma+\lim_{h\to\infty}\left(\ln\left(h\right)-\sum_{k=n+1}^{h}\frac{1}{k} \right) $$

Answer 1

We can rewrite $$ \gamma+\lim_{h\to\infty}\left(\ln\left(h\right)-\sum_{k=n+1}^{h}\frac{1}{k} \right)=\gamma+\lim_{h\to\infty}\left(\ln\left(h\right)-\left[\sum_{k=1}^{h}\frac{1}{k} -\sum_{k=1}^{n}\frac{1}{k}\right]\right). $$ $$ =\gamma+\lim_{h\to\infty}\left(\ln\left(h\right)-\sum_{k=1}^{h}\frac{1}{k}\right)+ \sum_{k=1}^{n}\frac{1}{k}=\gamma-\lim_{h\to\infty}\left(\underbrace{\sum_{k=1}^{h}\frac{1}{k}}_{H_h}-\ln\left(h\right)\right)+H_n $$ By definition $\gamma := \lim\limits_{h\to\infty}\left(H_h - \ln h\right)$, so we have $$\gamma-\gamma+H_n=H_n.$$