A misunderstanding concerning $\pi$ The very well-known expression
$$\frac {\pi} {4} = 1 - \frac {1} {3} + \frac {1} {5} - \frac {1} {7} + \cdots$$
puts me face to face with a contradictory position. Let
$$s_N = \sum_{k = 0}^{N} \frac {1} {4k + 1} - \sum_{k = 0}^{N} \frac {1} {4k + 3}.$$
Then it is obvious that
$$\frac {\pi} {4} = \lim_{N \to \infty} s_N.$$
By Euler-MacLaurin summation formula,
$$\sum_{k = 0}^{N} \frac {1} {4k + 1} = \int_{0}^{N} \frac {dx} {4x + 1} + \frac {1} {2} \left(1 + \frac {1} {4N + 1} \right) + o \left (\frac {1} {N^2} \right)$$
and
$$\sum_{k = 0}^{N} \frac {1} {4k + 3} = \int_{0}^{N} \frac {dx} {4x + 3} + \frac {1} {2} \left(\frac {1} {3} + \frac {1} {4N + 3} \right) + o \left (\frac {1} {N^2} \right).$$
We then have
$$s_N = \frac {1} {4} \log \left(3 - \frac {6} {4N + 3} \right) + \frac {1} {3} + \frac {1} {(4N + 1) (4N + 3)} + o \left (\frac {1} {N^2} \right)$$
and
$$\lim_{N \to \infty} s_N = \frac {\log 3} {4} + \frac {1} {3}.$$
But $\frac {\log 3} {4} + \frac {1} {3} \ne \frac {\pi} {4}$. How come? Where have I done the mistake?

The error surely lies in the $o\left({1\over N^2}\right)$ term(s).  Look at it this way:  Your use of Euler-Maclaurin would suggest 
$$\sum_{k=1}^N{1\over k}=\int_1^N{dx\over x}+{1\over2}\left(1+{1\over N}\right)+o\left({1\over N^2}\right)$$
as well, which would suggest
$$\sum_{k=1}^N{1\over k}-\log N\to{1\over2}$$
instead of $\gamma\approx0.5772$.
