# Inequality of harmonic number $\left| \sum_{i=1}^{n}\frac{1}{i}-\log{n}-\gamma \right| \leq \frac{10}{n}$

In my Number Theory textbook, it was quoted without proof that for all positive integers $$n$$, $$\left| \sum_{i=1}^{n}\frac{1}{i}-\log{n}-\gamma \right| \leq \frac{10}{n}$$ where $$\gamma = 0.577...$$ is the Euler–Mascheroni constant.
From my Calculus courses, I knew that $$\displaystyle\lim_{n\to\infty}\left(\sum_{i=1}^{n}\frac{1}{i}-\log{n}\right)=\gamma$$
However, I am not aware of the inequality above.
I have no ideas of how to prove it, and couldn't find sources about it.
Is it a well-known result? Is it approachable? Is there a name for it? Are there resources about it?
Alternatively, a proof is also very welcomed.
Thank you very much.

• I would have thought $\frac{1}{2n}$ might be tighter than $\frac{10}{n}$ Commented Nov 22, 2020 at 23:27
• See Euler-Maclaurin formula - en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula#Examples Commented Nov 22, 2020 at 23:29
• @Henry So, $\frac{1-\gamma}{n} \le \sum_{i=1}^n \frac{1}{i} - \log n - \gamma \le \frac{1}{2n}$. Also, $\lim_{n\to \infty} n(\sum_{i=1}^n \frac{1}{i} - \log n - \gamma) = \frac{1}{2}$. Commented Nov 24, 2020 at 5:21

Let $$u_n=\sum_{i=}^n \frac{1}{i}-\log n$$ and $$v_n=\sum_{i=1}^n\frac{1}{i}-\log(n+1)$$, then $$(u_n)$$ is a decreasing sequence and $$(v_n)$$ is a nondecreasing sequence. Since they both converge towards $$\gamma$$, we have $$v_n\leqslant\gamma\leqslant u_n$$ for all $$n$$, which means that $$0\leqslant\sum_{i=1}^n\frac{1}{i}-\log n-\gamma\leqslant u_n-v_n=\log\left(1+\frac{1}{n}\right)\leqslant\frac{1}{n}$$
Start with the definition of $$\gamma$$ as
$$\gamma=\lim_{N\to\infty}\left(\sum_{i=1}^N{1\over i}-\ln N\right)$$
\begin{align} \sum_{i=1}^n{1\over i}-\ln n-\gamma &=\lim_{N\to\infty}\left(\ln N-\ln n-\sum_{i=n+1}^N{1\over i} \right)\\ &=\lim_{N\to\infty}\left(\int_n^N{dt\over t}-\sum_{i=n+1}^N{1\over i}\right)\\ &=\lim_{N\to\infty}\sum_{i=n+1}^N\int_{i-1}^i\left({1\over t}-{1\over i}\right)dt\\ &\lt\lim_{N\to\infty}\sum_{i=n+1}^N\left({1\over i-1}-{1\over i}\right)\\ &={1\over n} \end{align}
Remark: A somewhat more careful estimate on the integral, using the concave nature of $$1/t$$, gives the inequality $$\lt1/(2n)$$, mentioned by Henry in comments.