Euler-Mascheroni constant [strategic proof] 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 $\gamma$.
 A: 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
$$
0.5<\gamma<0.692
$$
A: 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.
