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?
 A: 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 $\gamma\approx0.5772$.
A: such mistakes are sometimes difficult to track down! however, your idea is an interesting one. it leads to the following expansion:
$$
S=\sum_{k=0}^{\infty} \frac1{4k+1}-\frac1{4k+3} \\
=1 -\sum_{k=1}^{\infty} \frac1{4k-1}-\frac1{4k+1}
$$
hence
$$
1-S =\sum_{k=1}^{\infty} \frac1{4k-1}-\frac1{4k+1} \\
=2\sum_{k=1}^{\infty}\frac1{(4k)^2}\left(1-\frac1{(4k)^2}\right)^{-1} \\
=2\sum_{k=1}^{\infty}\sum_{n=1}^{\infty} \frac1{(4k)^{2n}} \\
=2\sum_{n=1}^{\infty}\frac1{4^{2n}}\sum_{k=1}^{\infty}\frac1{k^{2n}} \\
=2\sum_{n=1}^{\infty}\frac{\zeta(2n)}{2^{4n}}
$$
now, using the expression for the zeta function at even integers in terms of the Bernouilli numbers:
$$
2\zeta(2n)=(-1)^{n+1}\frac{(2\pi)^{2n}}{(2n)!}B_{2n}
$$
we obtain
$$
1-S = \sum_{k=1}^{\infty} (-1)^{n+1} \frac{B_{2n}}{(2n)!}\left(\frac{\pi}2\right)^{2n}
$$
and, rearranging, with $S=\frac{\pi}4$, we obtain
$$
\frac{\pi}4 = \sum_{n=0}^{\infty} \frac{B_{2n}}{(2n)!}\left(\frac{i\pi}2\right)^{2n}
$$
A: As pointed by others, you misused the Euler-Maclaurin formula.
Indeed, the next terms will involve some coefficients times powers of $1,\dfrac13,\dfrac1{4N+1}$ and $\dfrac1{4N+3}$. The constant terms do not vanish as $N\to\infty$.
