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I know two proofs about the approximation of Euler-Mascheroni constant $\gamma$ that are very technical. So I would like to know if someone has a strategic proof to show that $0.5<\gamma< 0.6$.

Let be $\gamma\in \mathbb{R}$ such that

$$\large\gamma= \lim_{n\to +\infty}\left[\left(1+\frac{1}{2}+\cdots+\frac{1}{n}\right)-\log{(n+1)}\right].$$

Show that $0.5<\gamma< 0.6$

P.S.: In my book, the author use $\log{(n+1)}$ in the limit definition of $\gamma$.

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  • $\begingroup$ Are you familiar with the digamma function? $\endgroup$ Jul 8, 2015 at 4:03
  • $\begingroup$ Hi @EemilWallin, I don't know the Digamma function, but it is never too late to learn math! :) $\endgroup$
    – Irddo
    Jul 8, 2015 at 4:13
  • $\begingroup$ It is related many things but mainly on this case, digamma(x) = $\gamma +$ the harmonic series upto $\frac{1}{x}.$ Another useful fact is that you can use this to evaluate the limit using an accuracy of $2$ digits, which will help your case. I can post an answer if you want to see a simplified proof of simply computing $\gamma$. It's not very strategic though. $\endgroup$ Jul 8, 2015 at 4:19
  • $\begingroup$ If you can, I would like see your solution! $\endgroup$
    – Irddo
    Jul 8, 2015 at 4:29

2 Answers 2

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Setting $n=1$ and $m=8$ into the following inequality involving harmonic numbers

$$ 2H_n-H_{n(n-1)}<\gamma<2H_m-H_{m^2} $$

gives

$$ 0.5<\gamma<0.692 $$

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Let $f(x)=\frac{1}{x}$ and $H_n=\sum_{k=1}^nf(k)$. For every $k$ take the segments $\overline{P_kP_{k+1/2}}$ and $\overline{P_{k+1/2}P_{k+1}}$, where $P_k=(k,f(k))$. Note that for every $k$ the sum of area of trapezes $Q_kP_kP_{k+1/2}Q_{k+1/2}$, and $Q_{k+1/2}P_{k+1/2}P_{k+1}Q_{k+1}$, where $Q_k=(0,k)$, is greater than the area below $f(x)$ between $k$ and $k+1$. This implies

$$\frac{f(k)+f(k+1)+2f(k+1/2)}{4}>\int_k^{k+1}f(x)dx=\ln(k+1)-\ln k.$$

So $I_n=\frac{1}{4}[\sum_{k=1}^nf(k)+\sum_{k=1}^nf(k+1)]+\frac{1}{2}\sum_{k=1}^nf(k+1/2)>\ln(n+1).$

But $\frac{1}{2}\sum_{k=1}^nf(k+1/2)=\sum_{k=1}^n\frac{1}{2k+1}=\frac{H_n}{2}-1+\sum_{k=1}^{n+1}\frac{1}{n+k}$. Therefore

$$I_n=H_n-\frac{5}{4}+\frac{f(n+1)}{4}+\sum_{k=1}^{n+1}\frac{1}{n+k},$$

so

$$H_n-\ln(n+1)>\frac{5}{4}-\frac{f(n+1)}{4}+\sum_{k=1}^{n+1}\frac{1}{n+k},$$

taking the limits and using that $\lim_{n\to +\infty}\frac{f(n+1)}{4}+\sum_{k=1}^{n+1}\frac{1}{n+k}=\ln 2,$ we have that

$$\gamma \geq 1,25-\ln 2>0,54.$$

I believe that the other inequality can be done by a similar approach.

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