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Evaluate of this sum: $$S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$$

Expand out the sum:

$$S=\prod_{k=1}^{1}\frac{2k}{k+2}+\prod_{k=1}^{2}\frac{2k}{k+3}+\prod_{k=1}^{3}\frac{2k}{k+4}+\cdots$$

$$S=\frac{2}{3}+\frac{2}{4}\cdot\frac{4}{5}+\frac{2}{5}\cdot\frac{4}{6}\cdot\frac{6}{7}+ \frac{2}{6}\cdot\frac{4}{7}\cdot\frac{6}{8}\cdot\frac{8}{9}+\cdots+\frac{2^nn!}{(2n)!\div (n+1)!}$$ I don't know what to do next...

Andronicus
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Sibawayh
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4 Answers4

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I think, your expansion of the product is incorrect.

I get:

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}=\sum_{j=1}^{\infty}\frac{2^j j!}{\frac{(2j+1)!}{(j+1)!}}=\sum_{j=1}^{\infty} 2^j \frac{j! (j+1)!}{(2j+1)!} =\sum_{j=1}^{\infty} \frac{2^j}{\binom{2j+1}{j}} = \frac{\pi}{2}$$

Edit:

If you want to know how to get to $\frac{\pi}{2}$ then look at the question and answer from: How to sum this series for π/2 directly?

thinkingeye
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  • Where did the $(j+1)!$ come from? – Andronicus Mar 18 '20 at 06:49
  • $\prod\limits_{k=1}^{j} k = 1 \cdot 2 \cdot 3 \cdot ... \cdot j = j!$ However the factorial is shifted by $j + 1$, so the product is starting by $j+2$. $\prod\limits_{k=1}^j \left[ (j+1)+k \right] = (j+1)+1 \cdot (j+1)+2 \cdot (j+1)+3 \cdot ... \cdot (j+1)+j = (j+2) \cdot (j+3) \cdot (j+4) \cdot ... \cdot (2j+1) = \frac{(2j+1)!}{(j+1)!}$ – thinkingeye Mar 18 '20 at 06:55
  • The expanded form of the second product does not seem right, but I see it now, thank you! – Andronicus Mar 18 '20 at 07:00
  • Yes, I forgot some brackets. $\prod\limits_{k=1}^{j} \left[ (j+1)+k \right] = \left[ (j+1)+1 \right] \cdot \left[ (j+1)+2 \right] \cdot ... \cdot \left[ (j+1)+j \right] = ...$ – thinkingeye Mar 18 '20 at 07:04
2

Correcting $(2n)!$ to $(2n+1)!$ in OP's nth term, we have

$$S=\sum_{n=1}^{\infty} \frac{2^n n! (n+1)!}{(2n+1)!} =\sum_{n=1}^{\infty}\frac{2^n n ~\Gamma(n) \Gamma(n+2)}{\Gamma(2n+2)}=\sum_{n=1}^{\infty}2^n n~B(n,n+2)= \sum_{n=1}^{\infty} 2n\int_{0}^{\pi/2} 2^n \sin^{2n-1} x ~\cos^{2n+3} xdx.$$
$$S=\int_{0}^{\pi/2} \sqrt{2}\cos^4 x \sum_{n=1}^{\infty} n~(\sqrt{2} \sin x \cos x)^{2n-1}$$ Using the infinite GP result that $\sum_{1}^{\infty} n ~z^{2n-1} =\frac{z}{(1-z^2)^2}$ We get $$S=\int_{0}^{\pi/2} \sqrt{2} \cos^4 x \frac{\sqrt{2} \sin x \cos x}{(1-2\sin^2 x\cos^2 x)^2} dx= \int_{0}^{\pi/2} \frac{2 \sin x \cos^5x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Next using $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx,$ we get $$S=\int_{0}^{\pi/2} \frac{2 \sin^5 x \cos x}{(1-2\sin^2 x \cos^2 x)^2} dx$$ Adding the last two integrals, we get So $$2S=\int_{0}^{\pi/2} \frac{2\sin x \cos x(\sin^4 x+ \cos^4 x)}{(1-2\sin^2 x \cos^2 x)^2}=\int_{0}^{\pi/2} \frac{2\sin x \cos x}{(1-2\sin^2 x \cos^2 x)}dx=\int_{0}^{\pi/2} \frac{4\sin 2x dx}{1+\cos^2 2x}$$ $$\implies 2S=8\int_{0}^{\pi/4}\frac{\sin 2x dx}{1+\cos^2 2x}=-4\int_{1}^{0}\frac{dt}{1+t^2}=\pi \implies S=\frac{\pi}{2}$$ Lastly we have used $\cos 2x=t.$

Z Ahmed
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The product can be simplified to:

$$\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}= \sum_{j=1}^{\infty} \frac{j!(j+1)! 2^j}{(2j+1)!}= \sum_{j=1}^{\infty} \frac{2^j}{\binom{2j+1}{j}}$$

Andronicus
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1

Amazing is to recognize some series.

Consider $$S=\sum_{j=1}^{\infty} \frac{j!\,(j+1)!\, 2^j}{(2j+1)!}x^{2j}$$ Let $x=y \sqrt 2$ to make $$S=\sum_{j=1}^{\infty}\frac{4^j\, j!\, (j+1)! }{(2 j+1)!}y^{2 j}=-\frac{y^2}{y^2-1}+\frac{1}{2 \left(y^2-1\right)}-\frac{\sin ^{-1}(y)}{2 \sqrt{1-y^2} \left(y^2-1\right) y}$$ Making $y=\frac 1 {\sqrt 2}$ gives the result.

If we expand the rhs as a Taylor series built around $y=\frac 1 {\sqrt 2}$ we have $$S=\frac{\pi }{2}+\sqrt{2} (4+\pi ) \left(y-\frac{1}{\sqrt{2}}\right)+O\left(\left(y-\frac{1}{\sqrt{2}}\right)^2 \right)$$ $$S=\frac{\pi }{2}+(4+\pi ) (x-1)+O\left((x-1)^2\right)$$

Claude Leibovici
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