Very slowly converging formulas for $\pi$ I'm aware that both Leibniz's formula and Willis' product formula for $\pi$ converges at a logarithmic rate.
In a French paper, this formula was given as: "The slowest and heaviest formula imaginable to access $\pi,$ developed to verify a hypothesis on the volume of the sphere"

$S_m = \frac{4}{2^{m+1}} + \frac{4}{2^m}\sum\limits_{n=1}^{2^m-1}\sqrt{1-\left(\frac{n}{2^m}\right)^2} \enspace ; \enspace S_m \rightarrow \pi \enspace ; \enspace m \rightarrow \infty$

How slowly does the above formula converge?  Is it slower than the more classical formulas?  Are there formulas that converge even slower than the logarithmic rate?  Are there any good examples?
 A: 
How slowly does the above formula converge?

Not so slow, actually. But that's because one takes a subsequence with exponentially growing indices from a sequence that converges slower. Let $f(x) = \sqrt{1 - x^2}$, and for a positive integer $k$, define
$$T_k = \frac{1}{2k} \sum_{n = 1}^k \Biggl(f\biggl(\frac{n-1}{k}\biggr) + f\biggl(\frac{n}{k}\biggr)\Biggr).$$
Then $T_k$ is a trapezium sum for the integral $\int_0^1 f(x)\,dx$, for an equidistant partition of $[0,1]$ into $k$ subintervals. And, as is readily seen, $S_m = 4T_{2^m}$.
Trapezium sums for integrals of sufficiently smooth functions don't converge too badly. If $g$ is twice continuously differentiable on $[a,b]$, then one has
$$\Biggl\lvert \int_a^b g(x)\,dx - \frac{1}{2k}\sum_{n = 1}^k \Biggl(g\biggl(\frac{n-1}{k}\biggr) + g\biggl(\frac{n}{k}\biggr)\Biggr)\Biggr\rvert \leqslant \frac{(b-a)^3}{12k^2}\lVert g''\rVert_{\infty}.$$
Now, here the integrand is smooth on $[0,1)$, but the graph of $f$ has a vertical tangent at $1$, whence $f''$ is unbounded and the above doesn't tell us how good the convergence actually is. And indeed the convergence is slower, but not catastrophically so. We have $\pi - 4T_k \in \Theta(k^{-3/2})$. (Rough outline of argument: The area of a circular segment with angle $0 < \varphi \leqslant \pi$ is $\frac{1}{2}(\varphi - \sin \varphi) \in \Theta(\varphi^3)$, and the angle of the segment between $\frac{n}{k}$ and $\frac{n+1}{k}$ is $\arccos \frac{n}{k} - \arccos \frac{n+1}{k} \in \Theta\bigl(\frac{1}{\sqrt{k}\,\sqrt{k-n}}\bigr)$. Since $\sum j^{-3/2}$ is convergent, we get an overall difference in $\Theta(k^{-3/2})$.)
Thus $T_k$ converges to $\pi/4$ faster than the Leibniz series or the Wallis product. But in terms of rate of convergence, all are logarithmic. And it's easy to modify the Leibniz series to get more accurate approximations, while that's not so easy for $T_k$.
Now, since $S_m = 4T_{2^m}$ we have
$$\pi - S_m \in \Theta\bigl(2^{-3m/2}\bigr),$$
and that means $S_m$ converges linearly to $\pi$, with rate of convergence $\frac{1}{\sqrt{8}}$. But unless $m$ is small, computing $S_m$ is rather cumbersome, so computing $\pi$ via $S_m$ is not a very attractive option.
Still, "The slowest and heaviest formula imaginable" seems to be an exaggeration.
A: Given a sequence $a_n$ converging to $x$, we can always make $b_n=a_{\lfloor \frac{n}{2}\rfloor}$, that will also converge to $x$, but more slowly.
