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I've found this $\pi$ formula:

$$ \pi =\lim_{n\to \infty }4\sum_{k=1}^{n} \frac{2 n^3 (1-2 k)^2 \left((k-1) k+n^2\right)}{\left(k^2+n^2\right)^2\left((k-1)^2+n^2\right)^2} $$

What is interesting is that the formula has a very simple geometric explanation. So it could have been found long time ago, long before the discovery of series. So this is also my question: there are so many formulas for $\pi$ that it is hard to say if it is a new one or not. Even if the convergence is slow, it's much better than the classic ArcTan series for $x=1$:

For example and $\pi=3.14159...$, then for $n=100$, the formula gives $3.14144$ against $3.15149$ for the ArcTan series. So 3 correct terms against 1 for the ArcTan series.

Does anybody know?

EDIT1: in fact, the same kind of formula can be used to compute $\text{arccos}(\cdot)$, $\text{arcsin}(\cdot)$ and $\text{arctan}(\cdot)$ formulas based on the area of the angle. See my other post Riemann sum formulas for $\text{acos}(x)$, $\text{asin}(x)$ and $\text{atan}(x)$.

EDIT2: in fact, both formulas given by Lucian are almost equivalent to the one I gave. Indeed, we can rewrite the formula above as: $$ \frac{\pi }{8}=\lim_{n\to \infty }\sum_{k=1}^{n} \frac{n^3 (2 k-1)^2 \left((k-1) k+n^2\right)}{\left(k^2+n^2\right)^2\left((k-1)^2+n^2\right)^2} $$ Now, in relation to large $n$, we can assume that $k-1\approx k$ and so: $$ \frac{\pi }{8}=\lim_{n\to \infty }\sum_{k=1}^{n} \frac{n^3 (2 k-1)^2 \left(k^2+n^2\right)}{\left(k^2+n^2\right)^2\left(k^2+n^2\right)^2}=\lim_{n\to \infty }\sum_{k=1}^{n} \frac{n^3 (2 k-1)^2 }{\left(k^2+n^2\right)^3} $$ and again assuming that $2k-1\approx 2k$: $$ \frac{\pi }{8}=\lim_{n\to \infty }\sum_{k=1}^{n} \frac{n^3 (2 k)^2 }{\left(k^2+n^2\right)^3}\text{ },\text{thus}\text{ }\frac{\pi }{32}= \lim_{n\to \infty }\sum_{k=1}^{n}\frac{n^3 k^2 }{\left(k^2+n^2\right)^3} $$ But with each simplification, the formula converges slower...

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    $\begingroup$ yes, i believe the arctan series is alternating and is conditionally convergent. So the arctan series for pi converges more slowly than others. $\endgroup$ – cgo Dec 3 '15 at 6:32
  • $\begingroup$ How did you even get to this expression? $\endgroup$ – Chinny84 Dec 3 '15 at 9:57
  • $\begingroup$ @Chinny84, in fact this comes from a surface calculation and more: I think that similar formulas can be used to calculate acos, asin and atan formulas. $\endgroup$ – Eddy Khemiri Dec 3 '15 at 10:09
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Precision such as the one employed in this expression is quite a bit overkill. One could simply

have asked for a proof of $~\displaystyle\lim_{n\to\infty}~\sum_{k=1}^n\dfrac{n^3~(2k-1)^2}{(n^2+k^2)^3}=\dfrac\pi8$ , which, come to think about it, looks

suspiciously similar to a simple Riemann sum...

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    $\begingroup$ Or, better yet, $~\displaystyle\lim_{n\to\infty}~\sum_{k=1}^n \dfrac{n^3~k^2}{(n^2+k^2)^3}=\dfrac\pi{32}$ $\endgroup$ – Lucian Dec 3 '15 at 13:37
  • $\begingroup$ the formula has nothing to see with this kind of formulas. In fact, the formula takes into account the surface of a quarter of a circle. $\endgroup$ – Eddy Khemiri Dec 3 '15 at 14:06
  • $\begingroup$ @EddyKhemiri: "has nothing to see": as said by Lucian, this is a Riemann sum, which is a way to express the computation of an integral, i.e. the area under a curve. In this case, $y=x^2/(1+x^2)^3$. The antiderivative features an $\arctan$ term which corresponds to the area of an eighth of a circle. $\endgroup$ – Yves Daoust Dec 4 '15 at 10:18
  • $\begingroup$ @Lucian, could provide a link to the proof for the formulas you gave? $\endgroup$ – Eddy Khemiri Dec 12 '15 at 8:37
  • $\begingroup$ @EddyKhemiri: As I said, they are simple Riemann sums. Try to rewrite them as $\dfrac1n\displaystyle\sum_{k=1}^nf\bigg(\dfrac kn\bigg)$ $\endgroup$ – Lucian Dec 12 '15 at 8:39

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