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I am trying to evaluate this sum: $$\sum_{n=0}^\infty(-1)^n \frac{\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}$$ Wolfram alpha is not able to get a closed form, but the approximation it gives resembles the value of $$\frac{1}{\sqrt{\pi}}(\pi-2)$$ and this can be proved using an integral and interchanging it with the sum. However, I am looking for a proof that doesn't require the use of integrals, if it exists.

By the way, this is just a special case of the most general
$$\sum_{n=0}^\infty(-1)^n \frac{\Gamma\left(\frac{n+a+1}{2}\right)}{\Gamma\left(\frac{n+a+2}{2}\right)}=\frac{2^{a-1}}{\sqrt\pi}\frac{a\Gamma^2\left(\frac{a}{2}\right)-2\Gamma^2\left(\frac{a+1}{2}\right)} {\Gamma(a)}$$ I wasn't able to prove this result, in any way, so If someone knows a proof to this or the previous result I'd like to read it.

EDIT:

Thank you all for your solutions. Now in order to try to solve the general $a$ sum, consider that it is sufficient to evaluate the following integral: $$\int_0^{\frac{\pi}{2}}\frac{\cos^a(x)}{1+\cos(x)}dx$$ Just expand $\frac{1}{1+\cos(x)}$ as a geometric series and then use Wallis integral to convince yourself of this fact.

I didn't include this fact before because I didn't want solutions that rely on this integral, since it seems much more doable then the sum, but now I feel it can help to get a full proof that otherwise would be very tough. I tried to evaluate the integral and got the infamous series, so if you have any ideas...

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    $\begingroup$ I'm working on a solution involving elementary antidifferentiation but no integrating in the way you describe. Is that permissible? $\endgroup$
    – user170231
    Commented Jun 12, 2023 at 19:30
  • $\begingroup$ Sure thank you @user170231 $\endgroup$
    – Zima
    Commented Jun 12, 2023 at 19:33
  • $\begingroup$ Are you looking for evaluating the integral? $\endgroup$
    – Kroki
    Commented Jun 13, 2023 at 0:22
  • $\begingroup$ Yes, please @ Youem $\endgroup$
    – Zima
    Commented Jun 13, 2023 at 5:37
  • $\begingroup$ The integral has the same value as the series? Am I missing something? $\endgroup$
    – Kroki
    Commented Jun 13, 2023 at 6:07

6 Answers 6

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Let $\alpha = a$, $$c_n = \frac{\Gamma\left(\frac{n+\alpha+1}{2}\right)}{\Gamma\left(\frac{n+\alpha+2}{2}\right)}$$ using the fact that $\Gamma(x+1) = x\Gamma(x)$, then $$c_{n+2} = \frac{n+\alpha+1}{n+\alpha+2}c_n$$

Let

$$f(x) = \sum_{n=0}^\infty (-1)^n c_n x^n$$

then,

$$f'(x) = \sum_{n=0}^\infty n(-1)^n c_n x^{n-1}$$

  • Start by proving that: $$\left(1-x^2\right) xf'(x) + \left(\alpha(1 - x^2) - x^2\right)f(x) = \beta - \gamma x$$ with \begin{align} \beta &= \alpha c_0\\ \gamma &= (\alpha + 1)c_1. \end{align} Indeed, \begin{align} \left(1-x^2\right) xf'(x) + \left(\alpha(1 - x^2) - x^2\right)f(x) &= \sum_{n=0}^{\infty} n(-1)^nc_nx^n - \sum_{n=0}^{\infty} n(-1)^nc_nx^{n+2} + \sum_{n=0}^{\infty} \alpha (-1)^n c_nx^n - \sum_{n=0}^{\infty} (\alpha+1) (-1)^n c_nx^{n+2}\\ &= \alpha c_0 -(\alpha+1)c_1 x + \sum_{n=0}^{\infty} (-1)^n \underbrace{\left((n+\alpha +2)c_{n+2} - (n+\alpha+1)c_n\right)}_{=0} x^{n+2}\\ &= \beta - \gamma x \end{align}

  • Let $g(x) = f(x)x^\alpha \sqrt{1 - x^2}$, then \begin{align} g'(x) &= f'(x) x^{\alpha}\sqrt{1-x^2} + \alpha f(x) x^{\alpha - 1}\sqrt{1-x^2} - f(x)x^{\alpha}\frac{x}{\sqrt{1-x^2}}\\ &= \frac{x^{\alpha - 1}}{\sqrt{1 - x^2}}\left(\left(1-x^2\right) xf'(x) + \left(\alpha(1 - x^2) - x^2\right)f(x)\right)\\ &=\frac{x^{\alpha - 1}}{\sqrt{1 - x^2}}\left(\beta - \gamma x\right) \end{align}

  • L'Hôpital's rule: since $g(1)=0$, \begin{align} f(1) &=\lim_{x\to 1} \frac{g(x)}{x^\alpha \sqrt{1-x^2}}\\ &=\lim_{x\to 1} \frac{g'(x)}{\displaystyle\alpha x^{\alpha-1}\sqrt{1-x^2} - x^{\alpha} \frac{x}{\sqrt{1-x^2}}}\\ &= \lim_{x\to 1} \frac{\displaystyle\left(\beta -\gamma x\right) \frac{x^{\alpha -1}}{\sqrt{1-x^2}}}{\displaystyle\alpha x^{\alpha-1}\sqrt{1-x^2} - x^{\alpha} \frac{x}{\sqrt{1-x^2}}}\\ &= \lim_{x\to 1} \frac{\displaystyle\left(\beta - \gamma x\right)}{\displaystyle\alpha \left(1-x^2\right) - x^2}\\ &= \gamma - \beta. \end{align}

  • To finish the proof you need to simplify $\gamma - \beta$ which is not difficult.

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  • $\begingroup$ I have only one question: when you apply L'Hopital rule, why do we assume the integral goes to $0$ when $x$ approaches $1$? @Youem $\endgroup$
    – Zima
    Commented Jun 12, 2023 at 23:32
  • $\begingroup$ You don't to write the integral. Just use the fact that $g(1)=0$ $\endgroup$
    – Kroki
    Commented Jun 12, 2023 at 23:40
  • $\begingroup$ This is truly a marvelous solution, such a new approach for me ,I really appreciated it, thank you! $\endgroup$
    – Zima
    Commented Jun 12, 2023 at 23:43
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    $\begingroup$ You are welcome! But the solution is not complete. Do you think you will manage the last step? $\endgroup$
    – Kroki
    Commented Jun 12, 2023 at 23:44
  • $\begingroup$ Sure, tomorrow I will complete it $\endgroup$
    – Zima
    Commented Jun 12, 2023 at 23:45
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Use beta function: $$2\int_{0}^{\pi/2} \sin^{2p-1}x~ \cos^{2q-1}x~ dx=\frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}.$$ $$S=\sum_{n=0}^\infty(-1)^n \frac{\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}=\frac{1}{\Gamma(1/2)}\sum_{n=0}^\infty(-1)^n \frac{\Gamma\left(\frac{n+2}{2}\right)\Gamma(\frac{1}{2})}{\Gamma\left(\frac{n+3}{2}\right)}=\frac{2}{\Gamma(1/2)}\int_{0}^{\pi/2}\sum_{n=0}^{\infty}(-1)^n \sin^{n+1}x~ dx.$$ Summing the infinite GP, we get $$S=\frac{2}{\sqrt{\pi}} \int_{0}^{\pi/2} \frac{\sin x ~dx}{1+\sin x}=\frac{2}{\sqrt{\pi}}\left[\frac{\pi}{2}-\int_{0}^{\pi/2} (\sec^2 x-\sec x \tan x)dx\right]=\frac{2}{\sqrt{\pi}}\left[\frac{\pi}{2}- \lim_{x\to \pi/2}\frac{\sin x-1}{\cos x}-1\right]=\frac{\pi-2}{\sqrt{\pi}}.$$

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I don't believe there is a purely algebraic proof to the more general sum you have proposed, but there is one that makes use of the hypergeometric function. We generalize the given sum slightly and we look for an explicit formula for the function

$$S(x,a)=\sum_{n=0}^\infty x^n\frac{\Gamma\left(\frac{n+a+1}{2}\right)}{\Gamma\left(\frac{n+a+2}{2}\right)}$$

This looks like Kummer's hypergeometric function but it contains $n/2$ instead of $n$ in the argument of the summand. We deal with this by splitting into even and odd terms:

$$S(x,a)=\sum_{m=0}^\infty\frac{x^{2m}}{m!}\frac{\Gamma(m+(a+1)/2)\Gamma(m+1)}{\Gamma(m+(a+2)/2)}+x\sum_{m=0}^\infty\frac{x^{2m}}{m!}\frac{\Gamma(m+(a+2)/2)\Gamma(m+1)}{\Gamma(m+(a+3)/2)}$$

In this form the sum can be easily written as a linear combination of hypergeometrics:

$$S(x,a)=\frac{\Gamma\left(\frac{a+1}{2} \right)}{\Gamma\left(\frac{a+2}{2} \right)} ~_2F_1\left(1,\frac{a+1}{2},\frac{a+2}{2},x^2\right)+x\frac{\Gamma\left(\frac{a+2}{2} \right)}{\Gamma\left(\frac{a+3}{2} \right)} ~_2F_1\left(1,\frac{a+2}{2},\frac{a+3}{2},x^2\right)$$

Each of the two terms above diverge individually at $x=-1$ but when added together they should produce a finite result. Using known series expansions around the regular singular point $x=-1$ Mathematica is able to produce the finite result

$$\lim_{x\to -1}S(x,a)=2\left(\frac{\Gamma\left(\frac{a}{2}+1\right)}{\Gamma\left(\frac{1+a}{2}\right)}-\frac{\Gamma\left(\frac{1+a}{2}\right)}{\Gamma\left(\frac{a}{2}\right)}\right)$$

(these series expansions can be readily derived from linear 2nd order ODE theory, but they are too lengthy to reproduce here). After some simplification and using Legendre's duplication formula yields the result

$$S(-1,a)=\frac{2^{a-1}}{\sqrt{\pi}}\frac{a\Gamma^2\left(\frac{a}{2}\right)-2\Gamma^2\left(\frac{1+a}{2}\right)}{\Gamma(a)}$$

This formula also reproduces the $a=1$ result from the method in the other answer.

Regarding the case $a=1$: This sum also has an interesting integral representation which allows one to compute the sum in terms of elementary functions for integer values of $a$. Note that the summand can be written in terms of the Beta function as follows:

$$S(x,a)=\frac{1}{\sqrt{\pi}}\sum_{n=0}^\infty x^n B\left(\frac{n+a+1}{2}, \frac{1}{2}\right)$$

Using the integral representation $B(x,y)=\int_0^1 dt~ t^{x-1}(1-t)^{y-1}$ one can exchange the order of summation and integration and perform the resulting geometric sum to obtain the result

$$S(x,a)=\frac{1}{\sqrt{\pi}}\int_0^1\frac{dy}{\sqrt{y}}\frac{(\sqrt{1-y})^{a-1}}{1-x\sqrt{1-y}}=\frac{2}{\sqrt{\pi}}\int_0^1 du\frac{u^a}{(1-xu)\sqrt{1-u^2}}$$

This integral is close to a hypergeometric function, but not quite. It is true however by the integral representation found here that

$$J(x,a)= \frac{2}{\sqrt{\pi}}\int_0^1 du\frac{u^a}{(1-x^2u^2)\sqrt{1-u^2}}=\frac{\Gamma\left(\frac{a+1}{2} \right)}{\Gamma\left(\frac{a+2}{2} \right)} ~_2F_1\left(1,\frac{a+1}{2},\frac{a+2}{2},x^2\right)$$

and also $J(x,a)+xJ(x,a+1)=S(x,a)$ , which rederives the result we obtained above. Now, if we specialize to integer $a=M\in \mathbb{Z}$, we note that we can easily derive a recursion relation:

$$xS(x,M+1)=S(x,M)-\frac{2}{\sqrt{\pi}}\int_0^1 \frac{u^M}{\sqrt{1-u^2}}du=S(x,M)-\frac{\Gamma(M/2+1/2)}{\Gamma(M/2+1)}$$

It suffices then to calculate $S(x,0)$. With the substitution $u=\sin t$ the integral becomes elementary

$$\frac{\sqrt{\pi}}{2}S(x,0)=\int_0^{\pi/2}\frac{dt}{1-x\sin t}=\frac{\pi+2\text{arcsin}(x)}{2\sqrt{1-x^2}}$$

which confirms the calculation of $S(x,1)$ done above.

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  • $\begingroup$ Thank you! Can you please provide some additional details/links related to the derivation of the two series expansion you mention? $\endgroup$
    – Zima
    Commented Jun 12, 2023 at 19:47
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    $\begingroup$ I added some details about the case of integer $a$. If I find a reference for what you asked I'll post it. $\endgroup$ Commented Jun 12, 2023 at 22:53
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It is possible to evaluate the integral \begin{equation*} I(a)=\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^{a}(x)}{1+\cos(x)}\,\mathrm{d}x \end{equation*} via a combination of integration by parts and the beta function. \begin{gather*} I(a)= \int_{0}^{\frac{\pi}{2}}\dfrac{\cos^{a}(x)(1-\cos(x))}{\sin^2(x)}\,\mathrm{d}x =\underbrace{\left[-\cot(x)\cos^{a}(x)(1-\cos(x))\right]^{\frac{\pi}{2}}_{0}}_{= \, 0}+\\[2ex] +\int_{0}^{\frac{\pi}{2}}\cot(x)\left(a\cos^{a-1}(x)(-\sin(x)(1-\cos(x))+\cos^{a}(x)\sin(x)\right)\,\mathrm{d}x =\\[2ex] =\int_{0}^{\frac{\pi}{2}}\left((a+1)\cos^{a+1}(x)-a\cos^{a}(x)\right)\,\mathrm{d}x. \end{gather*} However, via the substitution $t=\cos^2(x)$ and the beta function we get \begin{equation*} \int_{0}^{\frac{\pi}{2}}\cos^{a}(x)\,\mathrm{d}x = \dfrac{1}{2}\int_{0}^{1}t^{\frac{a+1}{2}-1}(1-t)^{\frac{1}{2}-1}\,\mathrm{d}t = \dfrac{1}{2}B\left(\frac{a+1}{2},\frac{1}{2}\right)= \dfrac{\Gamma(\frac{1}{2})\Gamma(\frac{a+1}{2})}{2\Gamma(\frac{a+2}{2})}. \end{equation*} See https://en.wikipedia.org/wiki/Beta_function

Finally \begin{gather*} I(a)=\dfrac{\sqrt{\pi}}{2}\left(\dfrac{(a+1)\Gamma(\frac{a+2}{2})}{\Gamma(\frac{a+3}{2})}- \dfrac{a\Gamma(\frac{a+1}{2})}{\Gamma(\frac{a+2}{2})}\right)=\\[2ex] =\dfrac{\sqrt{\pi}}{2}\left(\dfrac{(a+1)\frac{a}{2}\Gamma(\frac{a}{2})}{\frac{a+1}{2}\Gamma(\frac{a+1}{2})}- \dfrac{a\Gamma(\frac{a+1}{2})}{\frac{a}{2}\Gamma(\frac{a}{2})}\right)=\\[2ex] =\dfrac{\sqrt{\pi}}{2}\left(\dfrac{a\Gamma(\frac{a}{2})}{\Gamma(\frac{a+1}{2})}- \dfrac{2\Gamma(\frac{a+1}{2})}{\Gamma(\frac{a}{2})}\right)=\\[2ex] =\dfrac{\sqrt{\pi}}{2}\dfrac{a\Gamma^{2}(\frac{a}{2})-2\Gamma^{2}(\frac{a+1}{2})}{\Gamma(\frac{a+1}{2})\Gamma(\frac{a}{2})}=\\[2ex] =\dfrac{\sqrt{\pi}}{2}\dfrac{a\Gamma^{2}(\frac{a}{2})-2\Gamma^{2}(\frac{a+1}{2})}{\sqrt{\pi}2^{1-a}\Gamma(a)} \\[2ex] =\dfrac{\sqrt{\pi}}{2}\cdot\dfrac{2^{a-1}}{\sqrt{\pi}}\dfrac{a\Gamma^{2}(\frac{a}{2})-2\Gamma^{2}(\frac{a+1}{2})}{\Gamma(a)} . \end{gather*} The sum of the series $=\frac{2}{\sqrt{\pi}}I(a).$

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  • $\begingroup$ Great solution thank you $\endgroup$
    – Zima
    Commented Jun 14, 2023 at 10:39
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Let $$f(x)=\sum_{n=0}^\infty(-1)^n \frac{\Gamma\left(\frac{n+2}{2}\right)}{\Gamma\left(\frac{n+3}{2}\right)}x^n. $$ Mathematica is able to give $$ f(x)=\frac{-\pi+\pi\sqrt{1-x^2}+2\arcsin(x)}{\sqrt{\pi}x\sqrt{1-x^2}}. $$ So $$ f(1)=\lim_{x\to 1}\frac{-\pi+\pi\sqrt{1-x^2}+2\arcsin(x)}{\sqrt{\pi}x\sqrt{1-x^2}}=\frac{\pi-2}{\sqrt\pi}. $$ Update: Let $$ g(x)=\sum_{n=0}^\infty \frac{\Gamma(n+1)}{\Gamma\left(\frac{2n+3}{2}\right)}x^{2n}$$ and $$ h(x)=\sum_{n=1}^{\infty}\frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(n+1\right)}x^{2n-1}$$ and then $f(x)=g(x)-h(x)$. Mathematica is able to get these two functions.

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  • $\begingroup$ Interesting. Do you have any clue on how Mathematica arrives to that result? @xpaul $\endgroup$
    – Zima
    Commented Jun 12, 2023 at 19:35
  • $\begingroup$ I think Mathematica separates the sum into two parts. $\endgroup$
    – xpaul
    Commented Jun 12, 2023 at 20:58
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Rewrite gammas to factorials to binomial coefficients:

$$\begin{align*} \sum_{n=0}^\infty (-1)^n \frac{\Gamma\left(\frac n2+1\right)}{\Gamma\left(\frac n2 + \frac32\right)} &= \sum_{n=0}^\infty \left[\frac{\Gamma(n+1)}{\Gamma\left(n+\frac32\right)} - \frac{\Gamma\left(n+\frac32\right)}{\Gamma\left(n+2\right)}\right] \\ &= \sum_{n=0}^\infty \left[\frac{n!}{\left(n+\frac12\right)!} - \frac{\left(n+\frac12\right)!}{(n+1)!}\right] \\ &= \sum_{n=0}^\infty \left[\frac{2^{2n+1}}{\Gamma\left(\frac12\right) (2n+1) \binom{2n}n} - \frac{\Gamma\left(\frac12\right)}{2^{2n+1}} \left(2 - \frac1{n+1}\right) \binom{2n}n\right] \end{align*}$$

Following up on xpaul's hint, we find

$$\begin{align*} g(x) &= \sum_{n=0}^\infty \frac{\Gamma(n+1)}{\Gamma\left(n+\frac32\right)} x^{2n} \\ &= \frac2{\Gamma\left(\frac12\right)} \sum_{n=0}^\infty \frac{2^{2n}}{(2n+1) \binom{2n}n} x^{2n} \\ &= \frac{2\arcsin x}{\sqrt\pi\,x\sqrt{1-x^2}} \\[2ex] h(x) &= \sum_{n=0}^\infty \frac{\Gamma\left(n + \frac32\right)}{\Gamma(n+2)} x^{2n+1} \\ &= \frac{\Gamma\left(\frac12\right)}2 \sum_{n=0}^\infty \frac1{2^{2n}} \left(2-\frac1{n+1}\right) \binom{2n}n x^{2n+1} \\ &= \frac{\sqrt\pi}2 \left(\frac{2x}{\sqrt{1-x^2}} - \frac{2-2\sqrt{1-x^2}}x\right) \end{align*}$$

and so the sum is

$$\begin{align*} & \lim_{x\to1^-} \left[\frac{2\arcsin x}{\sqrt\pi\,x\sqrt{1-x^2}} - \frac{\sqrt\pi}2 \left(\frac{2x}{\sqrt{1-x^2}} - \frac{2-2\sqrt{1-x^2}}x\right)\right] \\ &= \frac1{\sqrt\pi} \lim_{x\to1^-} \frac{2\arcsin x - \pi\left(1-\sqrt{1-x^2}\right)}{x \sqrt{1-x^2}} \\ &= \frac1{\sqrt\pi} \lim_{x\to1^-} \frac{\frac{2-\pi x}{\sqrt{1-x^2}}}{\frac{1-2x^2}{\sqrt{1-x^2}}} = \boxed{\frac{\pi-2}{\sqrt\pi}} \end{align*}$$

where the last limit is obtained by applying l'Hôpital's rule.


We use the well-known Maclaurin expansions of $\arcsin$ and $\arcsin^2$ to get the closed forms for $g,h$. This involves several rounds of differentiating, multiplying by an appropriate power of $x$, then anti-differentiating and applying the fundamental theorem of calculus.

Solving for the series in $g$:

$$\begin{align*} \arcsin^2x &= \sum_{n=1}^\infty \frac{2^{2n-1}}{n^2 \binom{2n}n} x^{2n} \\ \implies \frac{2x\arcsin x}{\sqrt{1-x^2}} &= \sum_{n=1}^\infty \frac{2^{2n}}{n \binom{2n}n} x^{2n} \\ \implies \frac{x^2}{1-x^2} + \frac{x\arcsin x}{\left(1-x^2\right)^{3/2}} &= \sum_{n=1}^\infty \frac{2^{2n}}{\binom{2n}n} x^{2n} \\ \implies \frac{\arcsin x}{\sqrt{1-x^2}} &= \sum_{n=0}^\infty \frac{2^{2n}}{(2n+1) \binom{2n}n} x^{2n+1} \end{align*}$$

Solving for the series in $h$:

$$\begin{align*} \arcsin x &= \sum_{n=0}^\infty \frac{\binom{2n}n}{2^{2n}(2n+1)} x^{2n+1} \\ \implies \frac1{\sqrt{1-x^2}} &= \sum_{n=0}^\infty \frac{\binom{2n}n}{2^{2n}} x^{2n} \\[2ex] \frac 1{\sqrt{1-x}} &= \sum_{n=0}^\infty \frac{\binom{2n}n}{2^{2n}} x^{n} \\ \implies -2\sqrt{1-x} &= -2 + \sum_{n=0}^\infty \frac{\binom{2n}n}{2^{2n} (n+1)} x^{n+1} \\ \implies \frac{2-2\sqrt{1-x^2}}{x^2} &= \sum_{n=0}^\infty \frac{\binom{2n}n}{2^{2n} (n+1)} x^{2n} \end{align*}$$

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  • $\begingroup$ How do you go from $\frac{2x\arcsin x}{\sqrt{1-x^2}}$ to $\frac{x^2}{1-x^2} + \frac{x\arcsin x}{\left(1-x^2\right)^{3/2}}$? I can't see the trick $\endgroup$
    – Zima
    Commented Jun 14, 2023 at 21:14
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    $\begingroup$ In the previous line, differentiate then multiply by $x$. $\endgroup$
    – user170231
    Commented Jun 14, 2023 at 21:16
  • $\begingroup$ This is a neat solution with a great explanation, all the steps are clear, thank you! I have already accepted another answer, otherwise I would have accepted yours. $\endgroup$
    – Zima
    Commented Jun 14, 2023 at 21:19
  • $\begingroup$ No worries, I'm glad you found it helpful! $\endgroup$
    – user170231
    Commented Jun 14, 2023 at 21:25

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