# 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 $|x|<1$.

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.

• Very nice! This was one of the more joyful readings on this site so far. Whoever voted this answer down must have a very strange reason... – t.b. Apr 2 '11 at 8:27
• +1 for this wonderfully motivated derivation, Didier. I wonder how on earth you stumbled on Fourier: where do you teach, by the way :-) – Georges Elencwajg Apr 2 '11 at 8:34
• @elgeorges Ha! You noticed that... :-) – Did Apr 2 '11 at 8:44
• What an excellent integral to represent the binomial coefficients!! Thank you! I was actually looking for a bunch of these. The best one I found myself was $$\frac{2^{m+n}(-1)^{n}}{\pi i{}^{m+n}}\int_{-\infty}^{\infty}\frac{1}{(x-i)^{n+1}(x+i)^{m+1}}dx=\binom{m+n}{n}$$ but it is not as pretty. (My solution to this used a combinatorics identity to find the binomial power series, because the above identity is not the greatest for explaining, especially since the only way I know to prove it is contour integration. But I personally prefer integrals and swapping the order any day!!) – Eric Naslund Apr 3 '11 at 13:59
• @Eric Thanks. You are right that contour integration techniques are not far away from "my" formula for the binomial coefficient. – Did Apr 3 '11 at 15:34

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.

• -1: This seems uncalled for, or rather, unuseful...as much fun as it is to answer a question flat out, it's obvious that the OP is in homework mode. Please consider 'deleting' this answer like Eric did for his, and undelete if the OP fixes things. You're doing their homework for them which is uncool. – Mitch Apr 1 '11 at 18:51
• I'm surprised at people that downvoted a perfectly reasonable answer to this question. I'm just helping the OP along. – user02138 Apr 1 '11 at 18:58
• @Mitch - If you don't like the question, downvote it. If you think the answer given is badly written or contains errors, downvote it. But this is just not classy. – Brian Vandenberg Apr 1 '11 at 19:04
• @Brian: Thanks. I second your comment. – user02138 Apr 1 '11 at 19:05
• @Mitch - I suppose that's one interpretation. I see the stackexchange-style sites as a far more global resource. While the answer may only serve the OP by doing his homework for him, this isn't an IRC chat-room or newsgroup post where a small audience will get something out of the response. – Brian Vandenberg Apr 1 '11 at 21:46

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

• @Mitch - The negative value is talked about in the wikipedia link I'm about to link to. The non-integer values could be covered by the interpretation $n! = \Gamma(n+1)$ -- though I'll confess, I don't know if there are any technicalities I'm missing with that blanket statement about non-integer $n$. Wikipedia link to binomial coefficients: en.wikipedia.org/wiki/… – Brian Vandenberg Apr 1 '11 at 21:53
• Can you give a tiny bit more? What does ${- \frac{1}{2} \choose 2k}$ even mean? – Mitch Apr 1 '11 at 21:55
• @Mitch - ${-n \choose k} = (-1)^{k}{{k - (n + 1)} \choose k}$, for positive $n$. The $(-1)^{k}$ coefficient keeps the value of the quantity positive (provided $n > k$?). Ignoring sign, this is the number of ways to arrange $|k - (n + 1)|$ items taken $k$ at a time. The $\frac{1}{2}$ I don't have a great intuitive description for, sorry. – Brian Vandenberg Apr 1 '11 at 22:13
• ${\alpha \choose k} = \frac{\alpha (\alpha-1) \ldots (\alpha-k+1)}{k!}$. I call that the "generalized" binomial coefficient (I don't know if it is the standard term). You can find more here en.wikipedia.org/wiki/…. And here is some info on the binomial series en.wikipedia.org/wiki/Binomial_series. – alejopelaez Apr 1 '11 at 22:58
• Nice...that's where the odd numbers come in the numerator, and then the $2^k k!$ in the denominator. – Mitch Apr 2 '11 at 2:33

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

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