Summing the power series $\sum\limits_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{2n+1}\prod\limits_{k=1}^n\frac{2k-1}{2k} $ I'd like to determine the function corresponding to the following power series:
$$x + \sum_{n=1}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} \frac{x^{2n+1}}{2n+1},
$$
where $|x|<1$.
 A: The following is an attempt to actually do some mathematics with the question. Thus I try to avoid ready-made formulas drawn out of the hat and I want to describe step by step a path which leads to the solution as routinely as possible.
My first reaction when I read the formula to be explained is that I am not too happy with the denominators $2n+1$. The series is typeset with some ratios $$\frac{x^{2n+1}}{2n+1}$$ and I do not know if this is meant as a hint or not but these ratios are obvious primitives, so I feel like differentiating the series $A(x)$ the OP wants to compute. Note that I have no real justification for such a move at this point but one has to start somewhere, hasn't one? Thinking hard about the terms of degree $0$ and $1$ in $A(x)$, I finally convince myself that $A(0)=0$ and that
$$
A'(x)=\sum_{n=0}^\infty (-1)^n\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n} x^{2n}.
$$
Now, what is this strange looking coefficient of $x^{2n}$? This looks like a ratio of factorials, but not quite... After some fiddling around, I notice that I could use the even integers in the denominator to fill in the gaps between the odd integers in the numerator and that the resulting numerator would then be a perfect factorial. Seems like a good idea, but first I must write this rigorously, getting
$$
\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}=\frac{1\cdot2\cdot3\cdots(2n)}{2^2\cdot4^2\cdot6^2\cdots(2n)^2}=\frac{(2n)!}{2^{2n}(n!)^2}=\frac1{2^{2n}}{2n\choose n}.
$$
My next step is to change variable. Using the variable $z=-\displaystyle\frac{x^2}4$ instead of $x$ can only simplify things, right? So now, my aim is to understand the series
$$
A'(x)=\sum_{n=0}^\infty{2n\choose n}z^n.
$$
Ha! Binomial coefficients! That is easy, I just have to find a binomial somewhere. The expansion of $(1+u)^{2n}$ involves all the $2n$-choose-something, so I am after the term $u^n$ in $(1+u)^{2n}$. How could I keep this term and erase all the others? Well, a guy named Joseph Fourier already knew how to do that, so I will try to imitate him.
Old Joseph knew that if you integrate the function $s\mapsto\mathrm{e}^{2\mathrm{i}\pi ns}\mathrm{e}^{-2\mathrm{i}\pi ks}$ on $[0,1]$ for integers $n$ and $k$, you get zero except in one case: if $k=n$, and then you get $1$. So if I replace $u$ by $\mathrm{e}^{2\mathrm{i}\pi s}$ in $(1+u)^{2n}$ and if I integrate everything multiplied by $\mathrm{e}^{-2\mathrm{i}\pi ns}$ on $[0,1]$, I will collect the coefficient of $u^n$ alone. In other words,
$$
{2n\choose n}=\int_0^1(1+\mathrm{e}^{2\mathrm{i}\pi s})^{2n}\mathrm{e}^{-2\mathrm{i}\pi ns}\mathrm{d}s.
$$
Now I multiply this by $z^n$, I sum the results over $n$ and I assume that $|z|$ is small enough to allow me to interchange the order of the summation and the integral (I see that $|z|<\frac14$ will do). The result is
$$
A'(x)=\int_0^1\sum_{n=0}^\infty z^n(1+\mathrm{e}^{2\mathrm{i}\pi s})^{2n}\mathrm{e}^{-2\mathrm{i}\pi ns}\mathrm{d}s.
$$
The function I integrate is nothing but a geometric series, right? And the modulus of its ratio is at most $4|z|<1$, right? Here I am in known territory because I know how to sum a geometric series
$$
G(r)=\sum_{n=0}^{+\infty}r^n,
$$
for every complex number $r$ such that $|r|<1$ and I know this because somebody told me once how another guy named Euclid did it: he (basically) wrote the series as
$$
G(r)=1+r+r^2+r^3+\cdots=1+r(1+r+r^2+\cdots),
$$
and he noticed that the parenthesis was $G(r)$ again! In other words, $G(r)=1+rG(r)$, that is,
$$
G(r)=\frac1{1-r}.
$$
So I can compute the series inside the integral and this gets me a simpler expression of $A'(x)$, namely
$$
A'(x)=\int_0^1\frac{\mathrm{d}s}{1-z(1+\mathrm{e}^{2\mathrm{i}\pi s})^{2}\mathrm{e}^{-2\mathrm{i}\pi s}}.
$$
This is an integral of a rational function of sines and cosines, in this case
$$
A'(x)=\int_0^1\frac{\mathrm{d}s}{1-4z\cos^2(\pi s)}=\frac2\pi\int_0^{\pi/2}\frac{\mathrm{d}s}{1-4z\cos^2(s)}=\frac2\pi\int_0^{\pi/2}\frac{\mathrm{d}s}{1+x^2\cos^2(s)},
$$
where in the last equality I finally decided to come back to the $x$ variable. Since $\cos^2(s)$ and $\mathrm{d}s$ are invariant by the transformation $s\to s+\pi$, I am pretty sure the change of variables $t=\tan(s)$ will be successful. Let me verify this: $\mathrm{d}t=(1+t^2)\mathrm{d}s$ and $\cos^2(s)=1/(1+t^2)$, hence
$$
A'(x)=\frac2\pi\int_0^{+\infty}\frac{\mathrm{d}t}{1+t^2+x^2}
=\frac2\pi\int_0^{+\infty}\frac{\mathrm{d}t}{\sqrt{1+x^2}(1+t^2)},
$$
that is
$$
A'(x)
=\frac2\pi\frac1{\sqrt{1+x^2}}\left[\arctan(t)\right]_0^{+\infty}
=\frac1{\sqrt{1+x^2}}.
$$
Almost done! I remember that $A(0)=0$, hence
$$
A(x)=\int_0^x\frac{\mathrm{d}v}{\sqrt{1+v^2}}=\left[\mathrm{arcsinh}(v)\right]_0^x=\mathrm{arcsinh}(x)=\log(x+\sqrt{1+x^2}).
$$
Done.
A: Consider the following series,
\begin{align}
\arcsin x = \sum_{n \geq 0} \frac{(2n)!}{2^{2n} (n!)^{2}} \frac{x^{2n+1}}{2n+1}  = \sum_{n \geq 0} \frac{(2n-1)!!}{(2n)!!} \frac{x^{2n+1}}{2n+1},
\end{align}
which converges for $|x| \leq 1$ and where $\frac{(2n-1)!!}{(2n)!!} = \frac{1 \cdot 3 \cdots (2n-1)}{2 \cdot 4 \cdots 2n}$. This can be derived by integrating the series expansion of $(1 -x^2)^{-1/2}$ termwise and paying careful attention to questions of convergence. Observe that
\begin{align}
-i \arcsin i x = x + \sum_{n \geq 1} (-1)^n \frac{(2n-1)!!}{(2n)!!} \frac{ x^{2n + 1}}{2n+1},
\end{align}
which is the series in question.
A: Hint Consider the binomial series expansion for $(1+x^2)^{-\frac{1}{2}}$. You should arrive at something very similar to that series.
A: I am going to use a fairly well known generating series.  Its proof will be left until the end.

Lemma We have that $$\frac{1}{\sqrt{1-4x}}=\sum_{n=0}^\infty \binom{2n}{n} x^n$$ where $\binom{2n}{n}$ refers to the central binomial coefficient.  

Now to solve your problem: Define $f(x)$ by
$$f(x)=x+\sum_{n=1}^{\infty}(-1)^{n}\frac{1\cdot3\cdot5\cdots(2n-1)}{2\cdot4\cdot6\cdots2n}\frac{x^{2n+1}}{2n+1}.$$
Notice that 
$$f(x)=x+\sum_{n=1}^{\infty}(-1)^{n}\frac{(2n)!}{n!n!}\frac{1}{4^{n}}\frac{x^{2n+1}}{2n+1}$$
so that $$f^{'}(x)=1+\sum_{n=1}^{\infty}(-1)^{n}\frac{(2n)!}{n!n!}\frac{x^{2n}}{4^{n}}=\sum_{n=0}^{\infty}\binom{2n}{n}\left(\frac{-x^2}{4}\right)^{n}.$$
By the lemma, we see that
$$f^{'}(x)=\frac{1}{\sqrt{1+x^{2}}}.$$
To solve $f(x)=\int_0^x \frac{1}{\sqrt{1+x^{2}}}$, use the identity $\sinh^2(x)+1=\cosh^2 (x)$ and make the substitution $x=\sinh (u)$. (Notice the constant of integration must be zero from the original definition of $f(x)$.)  Then we find $$f(x)=\sinh^{-1}(x).$$
Proof of Lemma: We have the combinatorial identity $$\sum_{i=0}^{n}\binom{2i}{i}\binom{2(n-i)}{n-i}=4^{n}$$ which follows from finding two different ways to count the number of possible ways to choose a team from a group of size $2n$.  Then 
$$\left(\sum_{n=0}^\infty \binom{2n}{n} x^n\right)^2=\sum_{n=0}^\infty (4x)^n=\frac{1}{1-4x}.$$ 
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 0}^{\infty}\pars{-1}^{n}{x^{2n + 1} \over 2n + 1}
\prod_{k = 1}^{n}{2k - 1 \over 2k}} =
\sum_{n = 0}^{\infty}\pars{-1}^{n}{x^{2n + 1} \over 2n + 1}
\prod_{k = 1}^{n}{k - 1/2 \over k}
\\[5mm] = &\
\sum_{n = 0}^{\infty}\pars{-1}^{n}{x^{2n + 1} \over 2n + 1}
{\pars{1/2}^{\large\overline{n}} \over n!} =
\sum_{n = 0}^{\infty}\pars{-1}^{n}{x^{2n + 1} \over 2n + 1}
{\Gamma\pars{1/2 + n}/\Gamma\pars{1/2} \over n!}
\\[5mm] = &\
\sum_{n = 0}^{\infty}\pars{-1}^{n}{x^{2n + 1} \over 2n + 1}
{\pars{n - 1/2}! \over n!\pars{-1/2}!} =
\sum_{n = 0}^{\infty}\pars{-1}^{n}{x^{2n + 1} \over 2n + 1}
{n - 1/2 \choose n}
\\[5mm] = &\
\sum_{n = 0}^{\infty}\pars{-1}^{n}{x^{2n + 1} \over 2n + 1}
{-1/2 \choose n}\pars{-1}^{n} =
\sum_{n = 0}^{\infty}{-1/2 \choose n}x^{2n + 1}\int_{0}^{1}t^{2n}\,\dd t
\\[5mm] = &\
x\int_{0}^{1}\sum_{n = 0}^{\infty}
{-1/2 \choose n}\pars{x^{2}t^{2}}^{n}\,\dd t =
x\int_{0}^{1}{\dd t \over \root{1 + x^{2}t^{2}}}
\\[5mm] = &\
\int_{0}^{x}{\dd t \over \root{1 + t^{2}}}
\,\,\,\stackrel{t\ =\ \sinh\pars{\theta}}{=}\,\,\,
\int_{0}^{\mrm{arcsinh}\pars{x}}\dd\theta
\\[5mm] = &\
\mrm{arcsinh}\pars{x} =
\bbx{\ln\pars{x + \root{1 + x^{2}}}}
\end{align}
