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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 $\frac{\mathrm{arsinh}(a)}{\sqrt{1+a^2}}$ –  J. M. Aug 28 '10 at 2:48
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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. –  J. M. Aug 28 '10 at 2:52
    
...and here's the hypergeometric case you need: functions.wolfram.com/07.23.03.3098.01 –  J. M. Aug 28 '10 at 3:32
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3 Answers

up vote 17 down vote accepted

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. –  J. M. Aug 28 '10 at 4:40
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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). –  Qiaochu Yuan Aug 28 '10 at 3:15
    
Good thing you reminded Moron: The products appearing in the numerators and denominators are the "double factorials". –  J. M. Aug 28 '10 at 3:18
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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). –  Aryabhata Aug 28 '10 at 4:01
    
Hey is there any book which I could refer for getting more used to this type of summation. –  user9413 May 16 '11 at 15:46
    
@Chandru: I am not sure. I guess most real analysis books should have stuff like this. –  Aryabhata May 16 '11 at 17:19
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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] –  J. M. Aug 28 '10 at 4:13
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Never knew the pochhammer symbol. +1 just for that :-) –  Aryabhata Aug 28 '10 at 14:08
    
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. –  J. M. Aug 28 '10 at 14:13
    
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). –  J. M. Aug 28 '10 at 14:16
    
A differential equation is actually quite standard. No question though, it is elegant. I was going for something different. –  Aryabhata Aug 28 '10 at 14:18
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