From the equation for a circle, \(x^2+y^2=1\), we know its total area is \(\pi\), and can form the following integral for computing its area utilizing symmetry and focusing on a quarter circle. \[4\int_0^1\sqrt{1-x^2}dx\] First we need to turn it into a power series. \[(1-x^2)^{1/2}=\sum_{k=0}^\infty{1/2\choose k}(-1)^kx^{2k}\] Simplifying the binomial coefficient (also afterward, ensure the result matches the original series since \(2k-3<0\) when \(k=0,1\)) \[{1/2\choose k}={{1\over2}{-1\over2}{-3\over2}\ldots{-(2k-3)\over2}\over k!} ={(-1)^{k-1}(2k-3)!!\over2^kk!}\] \[={(-1)^{k-1}(2k-1)!!\over2^kk!(2k-1)} {(2k)!!\over2^kk!}={(-1)^{k-1}(2k)!\over4^k(k!)^2(2k-1)} ={2k\choose k}{(-1)^{k-1}\over4^k(2k-1)}\] Now evaluate the integral of the power series \[4\int_0^1\left[\sum_{k=0}^\infty{2k\choose k}{-x^{2k}\over4^k(2k-1)}\right]dx =4\sum_{k=0}^\infty{2k\choose k}{-1\over4^k(2k-1)(2k+1)}\] \[=2\sum_{k=0}^\infty{2k\choose k}{1\over4^k} \left({1\over1+2k}+{1\over1-2k}\right)\] \[=2\sum_{k=0}^\infty{2k\choose k}{1\over4^k(2k+1)} +2\sum_{k=0}^\infty{2k\choose k}{1\over4^k(1-2k)}\] So now we find that the integral evaluates into 2 summations. The first summation is actually the series expansion for \(\arcsin(1)\) so we know that the solution, \(\pi\), comes from there, and that this method does not give us a new series for approximating \(\pi\). But then we still have another summation which must be zero by this reasoning. We can show that it is zero, but that will not give us anything new for \(\pi\) approximations.
We would like to show \[\sum_{k=0}^\infty{2k\choose k}{1\over4^k(1-2k)}=0\] Notice how the \(1-2k\) part is negative except for the \(k=0\) term. Split up the summation this way and then we can negate remaining terms to have positive denominators. Reindex the series to start at \(k=0\) and do some algebra to make it look nicer. \[=1-\sum_{k=1}^\infty{2k\choose k}{1\over4^k(2k-1)} =1-\sum_{k=0}^\infty{2k+2\choose k+1}{1\over4^{k+1}(2k+1)}\] \[=1-{1\over4}\sum_{k=0}^\infty{2k\choose k}{(2k+2)(2k+1)\over (k+1)(k+1)}{1\over4^k(2k+1)} =1-{1\over2}\sum_{k=0}^\infty{2k\choose k}{1\over4^k(k+1)}\] From this, we could recognize a generating function appropriate for the summation. But for now, notice that it is similar to the power series for \(\arcsin\) which comes from this integral: \[\int_0^x{dt\over\sqrt{1-t^2}} =\sum_{k=0}^\infty{2k\choose k}{x^{2k+1}\over4^k(2k+1)}\] When we use a binomial expansion on this, the \(t^2\) gives us exponents of \(t^{2k}\) which integrate to \({x^{2k+1}\over2k+1}\). If it was \(t\) instead of \(t^2\), we would get exponents of \(t^k\) integrating to \({x^{k+1}\over k+1}\). We can turn it into a power series and evaluate the integral to check that we get exactly the summation we are looking for. \[\int_0^x{dt\over\sqrt{1-t}}=\int_0^x(1-t)^{-1/2}dt =\int_0^x\sum_{k=0}^\infty{-1/2\choose k}(-1)^kt^kdt\] \[=\sum_{k=0}^\infty{-1/2\choose k}{(-1)^kx^{k+1}\over k+1}\] From deriving the power series for \(\arcsin\) or by other means, we can show \[{-1/2\choose k}={2k\choose k}{(-1)^k\over4^k}\] So substituting this, we get \[\int_0^x{dt\over\sqrt{1-t}} =\sum_{k=0}^\infty{2k\choose k}{x^{k+1}\over4^k(k+1)}\] Now we use \(x=1\) so the summation identically matches what we were looking at before and show that it evaluates to \(2\). \[\int_0^1{dt\over\sqrt{1-t}}=\int_1^0(1-t)^{-1/2}d(1-t) =\left[{(1-t)^{1/2}\over1/2}\right]_1^0\] \[=2\left[(1-0)^{1/2}-(1-1)^{1/2}\right]=2\] Therefore, the strange summation that appeared turns out to be equal to zero. The integration of each \(\sqrt{1-x^2}\) and \({1\over\sqrt{1-x^2}}\) can give us \(\pi\), but it seems that \(\sqrt{1-x^2}\) only gives us the very slow series \(\arcsin(1)\) with further complications.