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How does one sum the series $$ S = a -\frac{2}{3}a^{3} + \frac{2 \cdot 4}{3 \cdot 5} a^{5} - \frac{ 2 \cdot 4 \cdot 6}{ 3 \cdot 5 \cdot 7}a^{7} + \cdots $$

This was asked to me by a high school student, and I am embarrassed that I couldn't solve it. Can anyone give me a hint?!

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    I'm getting $\frac{\mathrm{arsinh}(a)}{\sqrt{1+a^2}}$2010-08-28
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    As can be ascertained from the comment I gave, it's a special case of a hypergeometric function, since the coefficients multiplying the powers of a are expressible as ratios of gamma functions.2010-08-28
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    ...and here's the hypergeometric case you need: http://functions.wolfram.com/07.23.03.3098.012010-08-28

3 Answers 3

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HINT $\quad \:\;\;\rm (a^2+1) \: S' = 1 - a \: S \;\:$ by transmuting the coefficient recurrence to a differential equation.

$\rm\;\Rightarrow\; 1 = (a^2+1) \: S' + a \: S \; = \; f \: (f \; S)' \;\;$ for $\rm\;\; f = (a^2+1)^{1/2}$

$\rm\displaystyle\;\Rightarrow\; S = f^{-1} \int \; f^{-1} = \frac{\sinh^{-1}(a)}{(a^2+1)^{1/2}}$

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    Hmm, Frobenius. Nice! +1.2010-08-28
11

You can use the formula

$\displaystyle \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx} = \frac{2 \cdot 4 \cdot 6 \cdots 2k}{3\cdot 5 \cdots (2k+1)}$

This is called Wallis's product.

So we have $\displaystyle S(a) = \sum_{k=0}^{\infty} (-1)^k a^{2k+1} \int_{0}^{\frac{\pi}{2}}{\sin^{2k+1}(x) dx}$

Interchanging the sum and the integral

$\displaystyle S(a) = \int_{0}^{\frac{\pi}{2}}{\sum_{k=0}^{\infty}{(-1)^{k}(a\sin x)^{2k+1}} dx}$

The sum inside the integral is a geometric series of the form

$\displaystyle x - x^3 + x^5 - \cdots = x(1 - x^2 + x^4 - \cdots) = \frac{x}{1+x^2}$

Hence,

$\displaystyle S(a) = \int_{0}^{\frac{\pi}{2}}{\frac{a\sin x}{1 + (a\sin x)^2}}dx$

Now substitute $\displaystyle t = a \cos x$

The integral becomes

$\displaystyle \int_{0}^{a}{\frac{1}{1+a^2 - t^2}}dt = \frac{1}{2\sqrt{a^2+1}}\ln \left(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a} \right)$

Now $\displaystyle \ln \left(\frac{\sqrt{a^2+1}+a}{\sqrt{a^2+1}-a}\right) = \ln \left(\frac{\left(\sqrt{a^2+1}+a \right)^2}{\left(\sqrt{a^2+1}-a \right)\left(\sqrt{a^2+1}+a \right)}\right) = 2\ln \left(\sqrt{a^2+1}+a \right)$

So

$\displaystyle S(a) = \frac{1}{\sqrt{a^2+1}}\ln \left(\sqrt{a^2+1}+a \right) = \frac{\sinh^{-1}(a)}{\sqrt{a^2+1}}$

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    You add \displaystyle to the beginning of the TeX, and it's Wallis's product (at least the infinite version).2010-08-28
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    Good thing you reminded Moron: The products appearing in the numerators and denominators are the "double factorials".2010-08-28
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    @Qiaochu: Thanks for the tip about \displaystyle. About the product, the formula is exact (we don't need any infinite version) and is also called Wallis's product (at least I remember reading it that way).2010-08-28
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    Hey is there any book which I could refer for getting more used to this type of summation.2011-05-16
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    @Chandru: I am not sure. I guess most real analysis books should have stuff like this.2011-05-16
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    @Chandrasekhar: Sorry, I prefer to stay anonymous. Can't you just post the question here?2012-02-06
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    @Aryabhata: I can't do that. It's because, the problems a hard one and these guys will simply ask motivation and things like that, for which I am completely clueless. Anyway ok.2012-02-06
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    @Chandrasekhar: I don't think that should be a problem. You must have had some idea which you can post. For motivation, just reveal the source.2012-02-06
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    @Aryabhata: Well, then I shall say here itself: The question is this: $ 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{15} + \frac{1}{17} + \cdots \ \textrm{ad inf} = \frac{\pi}{8}\cdot\bigl(1+\sqrt{2}\bigr)$2012-02-06
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    @Chandrasekhar: Seems like you should be able to use this: http://en.wikipedia.org/wiki/Digamma_function2012-02-06
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    @Aryabhata: Source is "Integral Calculus for beginners" Joseph Edwards. Suna hai kya ye book.2012-02-06
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    @Chandrasekhar: Nope, nahin suna. Then there is probably an easier solution :-)2012-02-06
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    @Aryabhata: The author doesn't mention all this. He uses Gamma function, but not digamma. So i guess since he hasn't used it, it's premature of him to ask a problem like that. And that is a high-school book2012-02-06
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    @Aryabhata: This should work i think: $\int_{0}^{1}(1-x^{6})\cdot (1+x^{8}+x^{16} + \cdots \ \text{ad inf}) \ dx = \int_{0}^{1} \frac{(1-x^{6})}{1-x^{8}} \ dx$ is the required sum.2012-02-13
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    @Chandrasekhar: Well done, looks like it will.2012-02-13
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    @Aryabhata: I would like to practise these type of problems more. Could you suggest any problem book2012-02-13
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    @Chandrasekhar: I thought you already had one. Sorry, I don't know of any such problem books.2012-02-13
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    @Aryabhata: Oh, that book contains say 10 or 15 problems thats all2012-02-13
3

Making my comments more explicit:

Your sum of interest is

$\sum_{j=0}^\infty {(-1)^j \frac{(2j)!!}{(2j+1)!!} a^{2j+1}}$

where $(2j)!!=2\cdot 4\cdot 6\cdots (2j)$ and $(2j+1)!!=3\cdot 5\cdot 7\cdots (2j+1)$.

To simplify things a bit, we rearrange the series a bit to

$a\sum_{j=0}^\infty {\frac{(2j)!!}{(2j+1)!!}\left(-a^2\right)^j}$

The double factorials can be also expressed as

$(2j)!!=2^j j!=2^j (1)_j$

and

$(2j+1)!!=2^j \left(\frac32\right)_j$

where $(a)_j$ is a Pochhammer symbol.

Substitute both expressions into the series, and then note that the series now looks like a hypergeometric series. Now you can employ the formula here.

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    In *Mathematica*: `a Hypergeometric2F1[1, 1, 3/2, -a^2]`2010-08-28
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    Never knew the pochhammer symbol. +1 just for that :-)2010-08-28
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    Moron: Admittedly it's a bit highbrow; your solution is more elegant, but I think we should agree that Bill's is, most of all.2010-08-28
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    But to let you in on a dirty secret of mine: I just convert factorials, double factorials, binomial coefficients, Pochhammer symbols... into gamma functions, cancel what can be cancelled, and then just convert back to whatever function gives the neatest representation (See for instance that triple integral question I asked).2010-08-28
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    A differential equation is actually quite standard. No question though, it is elegant. I was going for something different.2010-08-28