Taylor's theorem II

Consequences of the series expansion of $\frac{1}{\sqrt{1-x}}$

Today we show the proof of the following series found on Twitter: \[\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n}x^n = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)\] \[ \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n^2} = \sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!n^2} = \frac{\pi^2}{6} -2\ln^2(2)\] In reality these series are just a transformation of a very simple series as we will show

Proof

We previously showed the proof of the following series expansion with the help of Pochhammer polynomials: \[ \frac{1}{\sqrt{1-x}} = \sum_{n=0}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^n = \sum_{n=0}^{\infty} \frac{\binom{2n}{n}}{2^{2n}}x^n ={}_{2}F_{1}\left(\frac{1}{2},1;1;x\right) \] This series, as simple as it is, will allow us to calculate other integrals and series much easier and this is one of those cases.
We will start subtracting $1$ from each side of the equation: \[ \frac{1}{\sqrt{1-x}}-1 = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^n\] Now divide by $x$ \[ \frac{1}{x\sqrt{1-x}}-\frac{1}{x} = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!} x^{n-1}\] Integrating from $0$ to $x$: \[ \int_{0}^{x}\frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!} \int_{0}^{x} x^{n-1}dx = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx \] The integral in the left hand side has the following antiderivative: \[ \int \frac{1}{x\sqrt{1-x}}-\frac{1}{x} dx = -2\ln\left(\sqrt{1-x}+1\right)+C\] Hence \[\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx = -2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2) = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)\] This is our first result: \[\boxed{\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)}\] If we divide by $x$ again we have: \[\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n-1}dx = \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} \] Integrating from $0$ to $1$: \[\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n^2(2n)!!} = \int_{0}^{1} \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} dx \] We can also write the left hand side with this form: \[\sum_{n=1}^{\infty}\frac{(2n-1)!!}{n^2(2n)!!} = \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{n^22^{2n}} \] This follow from the fact that: \[(2n-1)!! = \frac{(2n)!}{2^n n!}\] \[(2n)!! = 2^n n!\] \[ \Longrightarrow \frac{(2n-1)!!}{(2n)!!} = \frac{(2n)!}{2^{2n}n!^2} = \frac{1}{2^{2n}}\binom{2n}{n}\] Therefore: \[ \sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{n^22^{2n}} = \int_{0}^{1} \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} dx \tag{1}\] Integrating by parts: Let \[ du = \frac{1}{x} \quad v = -2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2) \] then \[ u = \ln(x) \quad dv = \frac{1}{(\sqrt{1-x}+1)\sqrt{1-x}} \] \begin{align*} \int_{0}^{1} \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} dx =& \ln(x)\left(-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)\right)\Big|_{0}^{1} - \int_{0}^{1} \frac{\ln(x)}{(\sqrt{1-x}+1)\sqrt{1-x}}\\ =& -\int_{0}^{1} \frac{\ln(x)}{(\sqrt{1-x}+1)\sqrt{1-x}}\\ =& -2\int_{0}^{1} \frac{\ln(1-w^2)}{w+1} dw\\ =& -2\int_{0}^{1} \frac{\ln\left((1+w)(1-w)\right)}{w+1} dw\\ =& -2\underbrace{\int_{0}^{1} \frac{\ln(1+w)}{w+1} dw}_{I_{1}} - 2\underbrace{\int_{0}^{1} \frac{\ln(1-w)}{w+1} dw}_{I_{2}} \tag{1} \end{align*} The integral $I_{1}$ is easily solvable given that \[ \int \frac{\ln(1+w)}{w+1} dw = \frac{1}{2} \ln^2(1+w) + C \] Hence \[ \displaystyle I_{1} = \frac{1}{2} \ln^2(2) \tag{3} \] For $I_{2}$: \begin{align*} I_{2}= \int_{0}^{1} \frac{\ln(1-w)}{w+1} dw =& \int_{0}^{1} \sum_{n=0}^{\infty}(-w)^n\ln(1-w) dw\\ =& \sum_{n=0}^{\infty}(-1)^n\int_{0}^{1} w^n\ln(1-w) dw\\ =& \sum_{n=0}^{\infty}(-1)^n\int_{0}^{1} w^n\left(\frac{d}{dt}\Big|_{t=0+}(1-w)^t\right) dw\\ =&\frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty}(-1)^n\int_{0}^{1} w^n(1-w)^t dw\\ =&\frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty}(-1)^nB(n+1,t+1)\\ =& \lim_{t \to 0+} \sum_{n=0}^{\infty}(-1)^n \left(\psi^{(0)}(t+1)-\psi^{(0)}(n+t+2)\right)B(n+1,t+)\\ =& \sum_{n=0}^{\infty}(-1)^n \left(\psi^{(0)}(1)-\psi^{(0)}(n+2)\right)B(n+1,1)\\ =& \sum_{n=0}^{\infty}(-1)^{n+1} \frac{H_{n+1}}{n+1}\\ =& \sum_{j=1}^{\infty}(-1)^{j} \frac{H_{j}}{j} \quad (j=n+1) \end{align*} Now recall the generating function for the harmonic numbers: \[ \sum_{n=1}^{\infty}H_{n}z^n = -\frac{\ln(1-z)}{1-z}\] Dividing by $z$ \[ \sum_{n=1}^{\infty}H_{n}z^{n-1} = -\frac{\ln(1-z)}{z(1-z)} = -\frac{\ln(1-z)}{z} -\frac{\ln(1-z)}{1-z}\] Integrating from $0$ to $-1$ \[ \sum_{n=1}^{\infty}\frac{H_{n}(-1)^{n}}{n} = -\int_{0}^{-1}\frac{\ln(1-z)}{z}dz -\int_{0}^{-1}\frac{\ln(1-z)}{1-z} = -\int_{0}^{1}\frac{\ln(1+z)}{z}dz +\int_{0}^{1}\frac{\ln(1+z)}{1+z}\] Hence \[ \sum_{n=1}^{\infty}\frac{H_{n}(-1)^{n}}{n} = -\int_{0}^{1}\frac{\ln(1+z)}{z}dz +I_{1} \tag{4} \] We just have to calculate $\displaystyle \int_{0}^{1}\frac{\ln(1+z)}{z}dz$ \begin{align*} \int_{0}^{1}\frac{\ln(1+z)}{z}dz =& \displaystyle \int_{0}^{1}\sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{k-1}}{k}dz\\ =& \sum_{k=1}^{\infty} (-1)^{k+1}\int_{0}^{1}\frac{z^{k-1}}{k}dz\\ =& \sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^2} \quad \textrm{(Dirichlet eta function)}\\ =& \eta(2)\\ =& \frac{\pi^2}{12} \end{align*} Therefore \[I_{2}= \sum_{n=1}^{\infty}\frac{H_{n}(-1)^{n}}{n} = -\frac{\pi^2}{12}+ \frac{1}{2} \ln^2(2) \tag{5} \] Hence, from (1),(2),(3),(4),(5) we can conclude: \[\int_{0}^{1} \frac{-2\ln\left(\sqrt{1-x}+1\right) + 2\ln(2)}{x} dx = -2I_{1} - 2I_{2} =\frac{\pi^2}{6} -2\ln^2(2)\] Therefore
\[ \boxed{\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n^2} = \sum_{n=1}^{\infty} \frac{(2n-1)!!}{(2n)!!n^2} = \frac{\pi^2}{6} -2\ln^2(2)} \] \[\boxed{\sum_{n=1}^{\infty} \frac{\binom{2n}{n}}{2^{2n}n}x^n = \sum_{n=1}^{\infty}\frac{(2n-1)!!}{n(2n)!!} x^{n}dx = 2\ln\left(\frac{2}{\sqrt{1-x}+1}\right)}\]