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} $

Question

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 $|x|<1$.

Answer 1

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.

Answer 2

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.

Answer 3

Hint Consider the binomial series expansion for $(1+x^2)^{-\frac{1}{2}}$. You should arrive at something very similar to that series.

Answer 4

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

$$\left(\sum_{n=0}^\infty \binom{2n}{n} x^n\right)^2=\sum_{n=0}^\infty (4x)^n=\frac{1}{1-4x}.$$

Answer 5

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}