TITLE: Showing that $\lim\limits_{n\to\infty}\sum^n_{k=1}\frac{1}{k}-\ln(n)=0.5772\ldots$
QUESTION [31 upvotes]: 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!

REPLY [44 votes]: Showing that
$\displaystyle\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\log(n)\right)$
exists
Note that
$$
\frac1{n+1}\le\int_n^{n+1}\frac1x\,\mathrm{d}x\le\frac1n\tag{1}
$$
Summing the right hand inequality shows that
$$
\sum_{k=1}^n\frac1k-\log(n+1)\tag{2}
$$
is increasing. Summing the left hand inequality shows that
$$
\sum_{k=1}^n\frac1k-\log(n)\tag{3}
$$
is decreasing.
Since $\displaystyle\lim_{n\to\infty}\log\left(\frac{n+1}{n}\right)=0$, $(2)$ and $(3)$ show that $\displaystyle\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1k-\log(n)\right)$ exists and is between $0$ and $1$.

Evaluating $\gamma$
Define the Harmonic Numbers
$$
H(n)=1+\frac12+\frac13+\frac14+\dots+\frac1n\tag{4}
$$
and the Alternating Harmonic Numbers
$$
AH(n)=1-\frac12+\frac13-\frac14+\dots+(-1)^{n-1}\frac1n\tag{5}
$$
and the Alternating Harmonic Tails
$$
AHT(n)=\frac1{n+1}-\frac1{n+2}+\frac1{n+3}-\frac1{n+4}+\dots\tag{6}
$$
Remember that
$$
\begin{align}
\log(2)&=1-\frac12+\frac13-\frac14+\frac15-\frac16+\dots\\
&=AH(n)+(-1)^nAHT(n)\tag{7}
\end{align}
$$
Notice that for $n\gt0$,
$$
\begin{align}
&H(2^n)-H(2^{n-1})\\
&=\left(1+\frac12+\frac13+\frac14+\dots+\frac1{2^n}\right)-\left(1+\frac12+\frac13+\frac14+\dots+\frac1{2^{n-1}}\right)\\
&=\left(1+\frac12+\frac13+\frac14+\dots+\frac1{2^n}\right)-\left(\frac22+\frac24+\frac26+\frac28+\dots+\frac2{2^n}\right)\\
&=\left(1-\frac12+\frac13-\frac14+\dots-\frac1{2^n}\right)\\
&=AH(2^n)\tag{8}
\end{align}
$$
Therefore,
$$
\begin{align}
H(2^n)
&=1+\sum_{k=1}^n\left(H(2^k)-H(2^{k-1})\right)\\
&=1+\sum_{k=1}^nAH(2^k)\tag{9}
\end{align}
$$
Because $\displaystyle\gamma=\lim_{n\to\infty}\Big(H(2^n)-\log(2^n)\Big)$, subtract $n\log(2)$ from $(9)$ and rearrange terms to get
$$
H(2^n)-\log(2^n)
=1+\sum_{k=1}^n\left(AH(2^k)-\log(2)\right)\tag{10}
$$
For $k\gt0$, $2^k$ is even, so we have from $(7)$
$$
\log(2)-AH(2^k)=AHT(2^k)\tag{11}
$$
Using $(10)$ and $(11)$, we get
$$
\gamma=1-\sum_{k=1}^\infty AHT(2^k)\tag{12}
$$
Expedite the computation of $(12)$ using the following integral,
$$
\begin{align}
AHT(m)
&=\sum_{j=1}^\infty\frac{(-1)^{j-1}}{m+j}\\
&=\int_0^1\frac{x^m}{1+x}\,\mathrm{d}x\\
&=\sum_{j=1}^\infty\frac1{j\binom{m+j}{j}2^j}\tag{13}
\end{align}
$$
where the last sum in $(13)$ is gotten by repeatedly integrating by parts. It converges far more quickly than the alternating harmonic tail, two lines above.
Using $(12)$ and $(13)$, it is not unreasonable to compute
$$
\gamma=0.57721566490153286060651209\tag{14}
$$
In fact, I used $(12)$ and $(13)$ to compute $10000$ digits of $\gamma$.

A Slightly Different Sum
Since $AHT(2^k-1)=2^{-k}-AHT(2^k)$, we can rewrite $(12)$ as
$$
\gamma=\sum_{k=1}^\infty AHT(2^k-1)\tag{15}
$$
and use $(13)$ and $(15)$ to get $(14)$, as well.

A Mathematica Implementation
Here is an implementation of the algorithm above to compute $d$ digits of $\gamma$.
AHT[m_,d_]:=Module[{p=10^(d+2), j=1, b=1, t, s=SetPrecision[1,d+4]-1}, 
  While[b=b*2(m+j)/j; (t=j b)<=p, s=s+1/t; j=j+1]; SetPrecision[s,d]]
EulerG[d_]:=Module[{n=Ceiling[10d/3]}, Sum[AHT[2^k-1,d], {k,1,n}]]