# Inequality involving Euler-Mascheroni constant

I would like to solve the following problem from 'Real Analysis and Foundations' by S. Krantz.

Let $$a_n = 1+\frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} - \log{n}$$ and $$\gamma$$ be the Euler-Mascheroni constant. Show that $$\vert a_n - \gamma \vert \le \frac{C}{n}$$ for some universal constant $$C>0$$.

I have no clue how to solve this problem. From searching various sites it seems that $$C=\frac{1}{2}$$ but I am not sure with this, too. Thanks in advance.

• Take a look here obtaining upper and lower bounds $1/(2n+2) < a_n - \gamma < 1/(2n)$.
– RRL
Commented Nov 7, 2018 at 6:31
• If a $C$ works, any larger $C$ would also work. There is no law dictating that you need to use this particular $C$. Commented Nov 7, 2018 at 6:34
• How to solve out this problem without integral? Commented Apr 15, 2021 at 2:03

From here we obtain the bounds

$$\frac{1}{2n+2} \leqslant a_n - \gamma \leqslant \frac{1}{2n}$$

Thus, if $$a_n - \gamma \leqslant \frac{C}{n}$$ then we must have $$\frac{C}{n} \geqslant \frac{1}{2n+2}$$ which implies for all $$n$$,

$$C\geqslant \frac{1}{2 + 2/n},$$

and $$C = \frac{1}{2}$$ is the smallest constant that works.

A self-contained approach. For any $$n\geq 1$$ we have $$\frac{1}{n}=\int_{0}^{+\infty}e^{-nt}\,dt$$ and $$\log(n)=\int_{0}^{+\infty}\frac{e^{-t}-e^{-nt}}{t}\,dt$$ by Frullani's theorem, hence $$a_n=H_n-\log(n)$$ can be represented as

$$\int_{0}^{+\infty}\frac{1-e^{-nt}}{e^t-1}-\frac{e^{-t}-e^{-nt}}{t}\,dt=\int_{0}^{+\infty}\left(\frac{1}{e^t-1}-\frac{1}{te^t}\right)\,dt-\int_{0}^{+\infty}\left(\frac{1}{e^t-1}-\frac{1}{t}\right)e^{-nt}\,dt$$ hence $$a_n-\gamma=\int_{0}^{+\infty}\left(\frac{1}{t}-\frac{1}{e^t-1}\right)e^{-nt}\,dt = \int_{0}^{+\infty}g(t) e^{-nt}. \tag{1}$$ We already know that the RHS goes to zero as $$n\to+\infty$$, and the positive function $$g(t)$$

• Is bounded by $$\frac{1}{2}e^{-t/6}$$ on the interval $$(0,6]$$
• Is bounded by $$\frac{1}{6}$$ on the interval $$(6,+\infty)$$

hence $$0\leq a_n-\gamma \leq \frac{3}{6n+1} \tag{2}$$ for any $$n\geq 2$$. By applying integration by parts to the RHS of $$(1)$$ you may similarly show that $$H_n = \log(n) + \gamma + \frac{1}{2n}-\frac{1-o(1)}{12n^2}.\tag{3}$$ The same can be proved through creative telescoping or through the Euler-Maclaurin summation formula.