Help me prove $\lim_{n→∞}\left(\sum_{k=1}^{n}\frac{1}{k} -\log n \right) >1/2$ I am a high school student.
I was able to prove that the limit of $\sum_{k=1}^{n}\frac{1}{k}-\log n$ converges. However, I can’t prove the convergence value is greater than $\frac{1}{2}$.
I tried a lot and was able to prove it is smaller than $1$. However, I can't greater than $\frac{1}{2}$.
 A: Copied from this answer:
Note that
$$
\begin{align}
\frac1n-\log\left(\frac{n+1}n\right)
&=\int_0^{1/n}\frac{t\,\mathrm{d}t}{1+t}\\
&\ge\int_0^{1/n}\frac{t}{1+\frac1n}\,\mathrm{d}t\\[3pt]
&=\frac1{2n(n+1)}
\end{align}
$$
Therefore,
$$
\begin{align}
\gamma
&=\sum_{n=1}^\infty\left(\frac1n-\log\left(\frac{n+1}n\right)\right)\\
&\ge\sum_{n=1}^\infty\frac1{2n(n+1)}\\[3pt]
&=\sum_{n=1}^\infty\frac12\left(\frac1n-\frac1{n+1}\right)\\[6pt]
&=\frac12
\end{align}
$$
A: We can apply the Euler-MacLaurin Formula, which says that
$$
\sum_{n=a}^b f(n) = \int_a^b f(x) dx + \frac{f(a)+f(b)}{2} + \sum_{k=1}^\infty \frac{B_{2k}}{(2k)!}(f^{(2k-1)}(b)-f^{(2k-1)}(a))+R
$$
In this case, $f(n) = 1/n$ and the above summation reduces to 
$$
\sum_{k=1}^n \frac{1}{n} = \log n + \frac{1}{2} + \frac{1}{2n} + Other_{terms}
$$
Show that the $Other_{terms}$ is positive. For this you need the definition of Bernouli numbers, the $(2k-1)$-th derivative of $1/x$ and the definition of $R$
A: Maybe you can show that $\sum_{k=1}^{\color{red}{n-1}}\frac{1}{k}-\log n$  converges to the same limit, but is increasing in $n.$ Then you can get arbitrarily good lower bounds by calculating this sequence for large enough $n.$
A: $$\gamma\stackrel{\text{def}}{=}\lim_{n\to +\infty}\left(H_n-\log n\right)=\sum_{n\geq 1}\left(\frac{1}{n}-\log\left(1+\frac{1}{n}\right)\right)\tag{A}$$
Now we may exploit the following facts to derive an integral representation for $\gamma$:
$$ \frac{1}{n}=\int_{0}^{+\infty}e^{-nx}\,dx,\qquad \log(n)=\int_{0}^{+\infty}\frac{e^{-x}-e^{-nx}}{x}\,dx\quad(\text{Frullani})\tag{B}$$
$$ \gamma = \int_{0}^{+\infty}\left(\frac{1}{e^x-1}-\frac{1}{x e^x}\right)\,dx =\int_{0}^{1}\left(\frac{1}{\log x}+\frac{1}{1-x}\right)\,dx$$
By the (generalized) Shafer-Fink inequality we may derive arbitrarily accurate uniform approximation for the $\arctan$ and $\text{arctanh}$ functions, hence for the logarithm. We have, for instance, $\log(x)\geq \frac{3(x^2-1)}{x^2+4x+1}$ for any $x\in(0,1)$. By plugging in such inequality in the previous integral representation we get
$$ \gamma \leq \frac{1+\log 2}{3} = 0.564382\ldots \tag{C1}$$
while by plugging in the improved inequality
$$\forall x\in(0,1),\qquad \log(x)\leq \frac{45(x^2-1)}{7+32(x+1)\sqrt{x}+x(12+7x)}$$
we get:
$$ \gamma \geq \frac{71-65\log 2}{45} = 0.576565\ldots\tag{C2} $$
This method can be used for producing even sharper approximations, see here.
