How to prove $\sum_{k=0}^n {n\choose k} \frac{(-1)^{k-1}}{2k-1} = 1+ \sum_{k=0}^{n-1} \frac{4^k (k!)^2}{(2k+1)!} = \frac{4^n (n!)^2}{(2n)!}$? In doing a calculation for my research I need a simple formula for the integral:
$$ I = 1 + \int_0^1 \frac{1-(1-x^2)^n}{x^2} dx$$
for natural $n$. Expanding the factor $(1-x^2)^n$ using the binomial theorem and integrating gives:
$$ I = \sum_{k=0}^n {n\choose k} \frac{(-1)^{k-1}}{2k-1}.$$
Alternatively, the integrand can be rewritten as a geometric sum:
$$ \frac{1-(1-x^2)^n}{x^2} = \sum_{k=0}^{n-1} (1-x^2)^k $$
and then each term in this sum can be integrated (and manipulated to give a beta function). The result gives:
$$ I =  1+ \sum_{k=0}^{n-1} \frac{4^k (k!)^2}{(2k+1)!}. $$
In order to match to the result I expected from this calculation I need it to be true that:
$$ I = \frac{4^n (n!)^2}{(2n)!} $$
and checking for values of $n < 100$ numerically these formulae do match and so I think this is correct. However, I don't know how to prove that either of my sum expressions or my initial integral expression for $I$ are equal to this final simple result. How can I show that $I$ is given by this final formula?
 A: If you are familiar with the Gaussian hypergeometric function
$$\int \frac{1-\left(1-x^2\right)^n}{x^2}\,dx=\frac 1 x\left(\,
   _2F_1\left(-\frac{1}{2},-n;\frac{1}{2};x^2\right)-1\right)$$
$$\int_0^1 \frac{1-\left(1-x^2\right)^n}{x^2}\,dx=\sqrt{\pi } \frac{\Gamma (n+1)}{\Gamma
   \left(n+\frac{1}{2}\right)}-1$$ So
$$I=\sqrt{\pi } \frac{\Gamma (n+1)}{\Gamma
   \left(n+\frac{1}{2}\right)}=\frac{4^n (n!)^2}{(2n)!}$$ If you need to be convinced, take logarithms and use Stirling approximation.
A: To compute $I$, just integrate by parts: $$I=1+\int_0^1\big(1-(1-x^2)^n\big)(-1/x)'\,dx=2n\int_0^1(1-x^2)^{n-1}\,dx,$$ and the integral is of beta type again: $$I=n\int_{-1}^1(1-x^2)^{n-1}\,dx\underset{x=1-2t}{\phantom{\big[}=\phantom{\big]}}2^{2n-1}n\mathrm{B}(n,n)=\texttt{(expected)}$$
A: A method without integrals.
\begin{align}
S_{n} &= \sum_{k=0}^{n} \binom{n}{k} \, \frac{(-1)^{k+1}}{2 k-1} \\
&= (-1) \, \sum_{k=0}^{n} \frac{(-n)_{k} }{k! \, (2k-1)} \\
&= (-1)^2 \, \sum_{k=0}^{n} \frac{(-1/2)_{k} \, (-n)_{k}}{k! \, (1/2)_{k}} \\
&= {}_{2}F_{1}\left(- \frac{1}{2}, \, -n; \, \frac{1}{2}; \, 1\right) \\
&= \frac{\Gamma\left(\frac{1}{2}\right) \, \Gamma(n+1)}{\Gamma(1) \, \Gamma\left(n + \frac{1}{2}\right) }  = \frac{n!}{\left(\frac{1}{2}\right)_{n}} \\
&= \frac{4^n}{\binom{2n}{n}} = \frac{4^n \, (n!)^2}{(2n)!}.
\end{align}
