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.

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

2 Answers 2


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.


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