Showing that $\lim\limits_{n\to\infty}\sum^n_{k=1}\frac{1}{k}-\ln(n)=0.5772\ldots$ How to show that  $$\lim_{n\to\infty}\left[\sum^n_{k=1}\frac{1}{k}-\ln(n)\right]=0.5772\ldots$$
No clue at all. Need help!  Appreciated!
 A: Abel's partial summation technique:
Let $A(n) = \displaystyle \sum_{k=1}^n a(k)$. We then have
\begin{align*}
\sum_{n=1}^{N} a(n) f(n) & = \sum_{n=1}^{N} f(n) (A(n)- A(n-1)) = \sum_{n=1}^{N} A(n) f(n) - \sum_{n=1}^{N} A(n-1) f(n)\\
& = \sum_{n=1}^{N} A(n)f(n) - \sum_{n=0}^{N-1} A(n) f(n+1)\\
& = A(N)f(N) - A(0) f(1) - \sum_{n=1}^{N-1} A(n) (f(n+1)-f(n))
\end{align*}
(The above is nothing but the discrete version of integration by parts).
$$\sum_{n=1}^{N} a(n) f(n) = \int_{1^-}^{N^+} f(t) d(A(t)) =  f(t) A(t) \rvert_{1^-}^{N^+} - \int_{1^-}^{N^+} A(t) f'(t) dt$$ (The second integral can be interpreted as a Riemann-Stieltjes integral.)

Now to the main proof. Consider the sum $\displaystyle \sum_{n \leq N} \frac1n$. Choose $a(n) = 1$ and $f(n) = \frac1n$. Note that we have $A(t) = \lfloor t \rfloor = t - \{t\}$. Hence, we get that
\begin{align*}
\sum_{n \leq N} \frac1n & = \left. \frac{t-\{t\}}t \right \rvert_{1^-}^{N^+} + \int_{1^-}^{N^+} \frac{(t-\{t\})}{t^2} dt\\
& = 1 + \int_{1^-}^{N^+} \frac{dt}t - \int_{1^-}^{N^+} \frac{\{t\}}{t^2} dt\\
& = 1 + \log (N) - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt + \int_{N^+}^{\infty} \frac{\{t\}}{t^2} dt\\
& = \left(1 - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt \right) + \log(N) + \int_{N^+}^{\infty} \frac{\{t\}}{t^2} dt
\end{align*}
Note that $\displaystyle \int_{N^+}^{\infty} \frac{\{t\}}{t^2} dt \leq \int_{N^+}^{\infty} \frac{1}{t^2} dt = \frac1N$.
Also note that by the same argument, we also have that $\displaystyle 0 < \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt < 1$ and hence $\displaystyle 0 < \left(1 - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt \right) < 1$. Denoting $\displaystyle \left(1 - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt \right) = \gamma$, we get the following result.
$$\sum_{n \leq N} \frac1n = \gamma + \log(N) + \mathcal{O} \left(\frac1N \right)$$
$\gamma \approx 0.5772\ldots$ and is called the Euler-Mascheroni constant. Hence, we have
$$\lim_{n \to \infty} \left(\sum_{k = 1}^n \dfrac1k - \log(n) \right) = \displaystyle \left(1 - \int_{1^-}^{\infty} \frac{\{t\}}{t^2} dt \right) = \gamma \approx 0.5772$$

We can easily see that $\gamma$ must be greater than $\dfrac12$ from the figure below. $\gamma$ is the area between the blue curve, which is $\displaystyle \dfrac1{\lfloor x \rfloor}$ and the red curve, which is $\dfrac1x$. The curve $\dfrac1x$ between $(n,n+1)$ lies below the line segment joining $\left(n,\dfrac1n \right)$ and $\left(n+1, \dfrac1{n+1} \right)$. Hence, \begin{align}
\gamma & = \text{Area between the blue and red curve}\\
& \geq \text{Area between the blue curve and the curve formed by the line segments }\\
& \text{joining $\left(n,\dfrac1n \right)$ and $\left(n+1, \dfrac1{n+1} \right)$}\\
& = \dfrac12 \times 1 \times \left(1-\dfrac12 \right) + \dfrac12 \times 1 \times \left(\dfrac12 - \dfrac13\right) + \dfrac12 \times 1 \times \left(\dfrac13 - \dfrac14\right) + \cdots\\
& = \dfrac12 \left(1 - \dfrac12 + \dfrac12 - \dfrac13 + \dfrac13 - \dfrac14 \pm\right) = \dfrac12 = 0.5
\end{align}

