Finding sum of

$\displaystyle 1+\frac{1\cdot 3}{6}+\frac{1\cdot 3 \cdot 5}{6\cdot 8}+\frac{1\cdot 3 \cdot 5 \cdot 7}{6 \cdot 8 \cdot 10}+\cdots \cdots$

Try: We can write sum as

$$ \mathcal{S} = 4\bigg[\frac{1}{4}+\frac{1\cdot 3}{4\cdot 6}+\frac{1\cdot 3 \cdot 5}{4\cdot 6 \cdot 8}+\cdots \cdots \cdots \bigg]$$

Now Let $$a_{n} = \prod^{n}_{k=1}(2k-1)=2^n\prod^{n}_{k=1}\bigg(k-\frac{1}{2}\bigg)$$

and $$b_{n} = 2^{-1}\prod^{n}_{k=1}(2k)=2^{n-1}\prod^{n}_{k=1}k$$

So $$\frac{a_{n}}{b_{n}} = 2\cdot \Gamma\left(n+\frac{1}{2}\right)\cdot \frac{1}{\Gamma\left(\frac{1}{2}\right)\cdot \Gamma(n+1)}$$

Above I have used $$\Gamma(x+n) = (x+n-1)(x+n-2)\cdot \cdots x \cdots \Gamma(n).$$

So $$\frac{a_{n}}{b_{n}} = 2\cdot \frac{\Gamma\left(n+\frac{1}{2}\right)\cdot \Gamma\left(\frac{1}{2}\right)}{\pi \cdot \Gamma(n+1)}= \frac{2}{\pi}\int^{1}_{0}x^{n-\frac{1}{2}}(1-x)^{-\frac{1}{2}}dx$$

So our sum is $$\mathcal{S} = \frac{8}{\pi} \sum^{\infty}_{n=1}\int^{1}_{0}x^n \cdot \frac{1}{\sqrt{x-x^2}}dx$$

So $$\mathcal{S} = \frac{8}{\pi}\int^{1}_{0}\frac{x}{(1-x)\sqrt{1-x^2}}dx$$

Put $x=\sin^2 \theta$ and $dx = 2 \sin \theta \cos \theta d\theta$ and changing limits

So we have $$\mathcal{S} = \frac{16}{\pi}\int^{\frac{\pi}{2}}_{0}(\sec^2 \theta -1)d \theta = $$

I did not understand where i am wrong.

and answer is $4$

Could some help me to explain it , thanks

  • $\begingroup$ It's unclear to me what your sum is, exactly. Could you add a few more terms in, please? $\endgroup$ – Matt Sep 18 '18 at 10:56
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    $\begingroup$ Is the term $\dfrac{1\cdot 3\cdot 5\cdot 7}{4\cdot 6\cdot 8\cdot 8}$ supposed to be $\dfrac{1\cdot 3\cdot 5\cdot 7}{6\cdot 8\cdot 10\cdot 12}$? $\endgroup$ – Batominovski Sep 18 '18 at 10:59
  • $\begingroup$ I also think $b_n$ is supposed to be $$2^n\,\prod\limits_{k=1}^{{\color{red}{n+1}}}\,k\,.$$ $\endgroup$ – Batominovski Sep 18 '18 at 11:01
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    $\begingroup$ Some questions involving similar series : math.stackexchange.com/questions/1091479/… and math.stackexchange.com/questions/1652349/… $\endgroup$ – Arnaud D. Sep 18 '18 at 11:06
  • $\begingroup$ So, your question is: $$ 1+\sum_{n=1}^{\infty}\prod_{k=1}^{n}\frac{2k+1}{2k+4}=4 \implies \sum_{n=1}^{\infty}\prod_{k=1}^{n}\frac{2k+1}{2k+4}=3 $$ $\endgroup$ – Hazem Orabi Sep 18 '18 at 11:12

You almost got everything right, and the only problem you have is a minor error when you define $b_n$. You should have $$b_n=2^n\,(n+1)!=2^n\,\Gamma(n+2)\text{ for }n=1,2,3,\ldots\,.$$ Thus, for each $n=1,2,3,\ldots$, $$\begin{align}\frac{a_n}{b_n}&=\frac{\Gamma\left(n+\frac12\right)}{\Gamma\left(\frac12\right)\,\Gamma(n+2)}=\frac{2}{\pi}\,\left(\frac{\Gamma\left(n+\frac12\right)\,\Gamma\left(\frac32\right)}{\Gamma(n+2)}\right)\\&=\frac{2}{\pi}\,\text{B}\left(n+\frac12,\frac32\right)\,,\end{align}$$ where $\Gamma$ and $\text{B}$ are the usual gamma and beta functions, respectively. Hence, $$\frac{a_n}{b_n}=\frac{2}{\pi}\,\int_0^1\,x^{n-\frac12}\,(1-x)^{\frac12}\,\text{d}x\,,$$ so $$\begin{align}\sum_{n=1}^\infty\,\frac{a_n}{b_n}&=\frac{2}{\pi}\,\int_0^1\,\frac{x^{\frac12}}{1-x}\,(1-x)^{\frac12}\,\text{d}x \\&=\frac{2}{\pi}\,\int_0^1\,x^{\frac12}\,(1-x)^{-\frac12}\,\text{d}x\,.\end{align}$$ That is, with $u:=x^{\frac12}$, we obtain $$\begin{align}\sum_{n=1}^\infty\,\frac{a_n}{b_n}&=\frac{4}{\pi}\,\int_0^1\,\frac{u^2}{\sqrt{1-u^2}}\,\text{d}u\\&=\frac{2}{\pi}\,\left(\text{arcsin}(u)-u\,\sqrt{1-u^2}\right)\Big|_{u=0}^{u=1}\,.\end{align}$$ Ergo, $$\sum_{n=1}^\infty\,\frac{a_n}{b_n}=1\,,$$ whence $$1+\frac{1\cdot 3}{6}+\frac{1\cdot 3\cdot 5}{6\cdot 8}+\frac{1\cdot3\cdot 5\cdot 7}{6\cdot 8\cdot 10}+\ldots=4\,\sum_{n=1}^\infty\,\frac{a_n}{b_n}=4\,.$$

In fact, one can show that $$f(z):=\sum_{n=0}^\infty\,\prod_{k=1}^n\,\left(\frac{k-\frac32}{k}\right)\,z^n=(1-z)^{\frac12}$$ for all $z\in\mathbb{C}$ with $|z|\leq 1$. The requested sum satisfies $$S=8\,\Biggl(1-\frac12-f(1)\Biggr)=4\,.$$

  • $\begingroup$ Thanks Batominovski for marvelous solution. $\endgroup$ – DXT Sep 18 '18 at 11:40

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