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

Question

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.

Answer 1

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 $$ 0\leqslant\sum_{i=1}^n\frac{1}{i}-\log n-\gamma\leqslant u_n-v_n=\log\left(1+\frac{1}{n}\right)\leqslant\frac{1}{n} $$

Answer 2

Start with the definition of $\gamma$ as

$$\gamma=\lim_{N\to\infty}\left(\sum_{i=1}^N{1\over i}-\ln N\right)$$

so

$$\begin{align} \sum_{i=1}^n{1\over i}-\ln n-\gamma &=\lim_{N\to\infty}\left(\ln N-\ln n-\sum_{i=n+1}^N{1\over i} \right)\\ &=\lim_{N\to\infty}\left(\int_n^N{dt\over t}-\sum_{i=n+1}^N{1\over i}\right)\\ &=\lim_{N\to\infty}\sum_{i=n+1}^N\int_{i-1}^i\left({1\over t}-{1\over i}\right)dt\\ &\lt\lim_{N\to\infty}\sum_{i=n+1}^N\left({1\over i-1}-{1\over i}\right)\\ &={1\over n} \end{align}$$

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.