TITLE: Euler's Constant: The asymptotic behavior of $\left(\sum\limits_{j=1}^{N} \frac{1}{j}\right) - \log(N)$
QUESTION [15 upvotes]: I want to show that there exists a constant $\gamma\in\mathbb{R}$ such that 
$$
\sum_{j=1}^N \frac1{j} = \log(N)+\gamma+O(1/N).
$$ 
I know how to prove that the Euler-Mascheroni constant exists (which I believe $\gamma$ to be), but I am having trouble with the big-$O$ notation and the subsequent bounding. I've considered 
$$
\left|\left(\sum_{j=1}^N \frac1{j}\right) - \log(N)-\gamma\right|\le |K/N|
$$ for some $K$, and I was approaching this by trying to show the that the left side of the inequality decays faster, but so far am stuck. Any advice for this type of problem, or analogous ones, would be appreciated. 
Thanks!

REPLY [3 votes]: By the MVT you have 
$$ \frac{\ln(k+1)- \ln(k)}{k+1-k}= \frac{1}{c_k} \,.$$
for some $k < c_k < k+1$.
Thus 
$$\frac{1}{k+1} < \ln(k+1)- \ln(k) < \frac{1}{k} \,.$$
From here, it is trivial to prove that 
$$a_n=1+\frac{1}{2}+..+\frac{1}{n} -\ln (n) $$
is decreasing to $\gamma$
while
$$b_n =1+\frac{1}{2}+..+\frac{1}{n} -\ln (n+1) $$
is increasing to $\gamma$.
It follows immediately that 
$$|a_n-\gamma| < |a_n -b_n| \,.$$