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The problem I have here today is the following; $$\frac{1}{4}+\frac{1\cdot3}{4\cdot6}+\frac{1\cdot3\cdot5}{4\cdot6\cdot8}+\cdots$$ the problem is exactly phrased like this (I can't say that the $\infty$ sign is a bit unnecessary at the end),

My Attempts

We can generalize this sum by noticing that each time indices get greater the denominator and numerators are multiplied by $n+2$ for each $n$ either in numerator or denominator, we take $\dfrac{1}{4}$ out of the sum first, so the sum is equal to $$\dfrac{1}{4}+\sum_{k=4} \frac{(k-3)(k-1)}{k(k+3)}$$ then we open up the brackets and we get; ....

Then I was a little stuck here because when I opened up these brackets and try to get the partitions of the sum one of them was logical $\displaystyle\sum\frac{3}{k(k+3)}=\sum\frac{1}{k}-\frac{1}{k+3}$. I couldn't carry it out longer. What do you suggest?

Is this a problem way above elementary solutions?

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    $\begingroup$ This is basically the same question without the factor of $\frac14$, you may find it helpful to read through some answers there. $\endgroup$
    – John Doe
    Oct 8, 2017 at 16:36
  • $\begingroup$ Thank you, I will take a look:) $\endgroup$ Oct 8, 2017 at 16:43
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    $\begingroup$ I don't think you have the general term correct. The general term is $a_0=\frac{1}{4}$ and $a_{n+1}=a_n\cdot \frac{2n+3}{2(n+3)}$. So the general term is $$a_n=\frac{1}{4}\prod_{k=1}^{n}\left(1-\frac{3}{2(k+3)}\right).$$ $\endgroup$ Oct 8, 2017 at 16:49
  • $\begingroup$ Idk why people like to put $\infty$ after $\dots$? Doesn't the $\dots$ cover the fact that it goes on forever? Adding an $\infty$ to the end usually causes confusion, for example, it might indicate that you want the $n$th term to approach $\infty$. $\endgroup$ Oct 8, 2017 at 17:24
  • $\begingroup$ @SimplyBeautifulArt That baffled me to, the forum I got this question had this $\infty$ sign so I didn't change it, I thought perhaps there was something more to it $\endgroup$ Oct 8, 2017 at 17:56

3 Answers 3

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You seem to be looking for a closed form for $$ \sum_{n\geq 0}\frac{(2n+1)!!}{2^{n+1}(n+2)!}=\sum_{n\geq 0}\frac{(2n+1)!}{2^{2n+1}(n+2)!n!}=\sum_{n\geq 0}\frac{(2n+2)!}{2^{2n+2}(n+2)!(n+1)!}$$ which is a telescopic series in disguise.
Once you recognize that, it is pretty clear that the series equals $\large 1$.


I realize there is an interesting by-product. The above series can be written as $$ \frac{2}{\pi}\sum_{n\geq 0}\frac{1}{n+2}\int_{0}^{\pi/2}\left(\sin\theta \right)^{2n+2}\,d\theta $$ hence the hidden telescopic structure provides a simple proof of the well-known identity: $$ \int_{0}^{\pi/2} \log\cos\theta\,d\theta = -\frac{\pi\log 2}{2}.$$

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    $\begingroup$ [+1] I applaude. $\endgroup$
    – Jean Marie
    Oct 8, 2017 at 17:08
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    $\begingroup$ Very insightful (+1). Can you please explain how you spot patterns like these and rearrange them? $\endgroup$ Oct 8, 2017 at 17:27
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    $\begingroup$ @hypergeometric: I was approaching the given problem through the second approach (brute-force conversion of a series into an integral), when I realized it was trivial by telescoping. After that, I realized my first attempt, together with the most elementary one, leads to a non-trivial consequence. $\endgroup$ Oct 8, 2017 at 17:42
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    $\begingroup$ @hypergeometric Also, Jack does this every day all the time, and I guess practice makes you better $\endgroup$ Oct 8, 2017 at 17:43
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    $\begingroup$ Absolutely incredible, thank you:)) $\endgroup$ Oct 8, 2017 at 17:54
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The solution below is inspired by this other solution here.

Note that

$$\begin{align} f(r)&=\frac {1\cdot 3\cdot 5\cdot\cdots \cdot(2r+1)}{4\cdot 6\cdot 8\cdot \cdots \cdot(2r+4)}\\ &=2\cdot \underbrace{\boxed{\frac {1\cdot 3\cdot 5\cdot\cdots \cdot(2r+1)}{2\cdot 4\cdot 6\cdot\cdots \cdot(2r+2)}}}_{A_r}\cdot \frac 1{2r+4}\\ &=2\ A_r\ \left(1-\frac {2r+3}{2r+4}\right)\\ &=2\left(A_r-A_{r+1}\right)\\ \frac 14+\frac {1\cdot 3}{4\cdot 6}+\frac {1\cdot 3\cdot 5}{4\cdot 6\cdot 8}+\cdots&=\sum_{r=0}^\infty f(r)\\ &=2\sum_{r=0}^\infty A_r-A_{r+1}\\ &=2\left(A_0-\lim_{r\to\infty}A_{r+1}\right)\\ &=2\left(\frac 12-0\right)\\ &=\color{red}1\end{align}$$


See also this for the limit of $A_{r+1}$.

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  • $\begingroup$ Nicely similar to my argument, but purely algebraic. Essentially, $f(r)=P(X=r+1)$, and you reduce it to $P(X>r)-P(X>r+1)$. $\endgroup$ Oct 8, 2017 at 19:02
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Amusingly, since the sum is $1$, you can actually write this as a question about a random number.

For each $n$, we flip an unfair coin $C_n$ with heads having probability of $\frac{1}{2(n+1)}$.

Let $X$ be the random variable which is $n$ if $C_n$ came up heads and for each $i<n$, $C_i$ came up tails.

Since $\prod_{k=1}^{n}\left(1-\frac{1}{2(k+1)}\right)\to 0$ as $n\to\infty$, you get $P(X<\infty)=1$.

Then it turns out that $$P(X=n)=\prod_{k=1}^{n}\frac{2k-1}{2k+2}$$

Proof: $$\begin{align}P(X=n)&=P(C_n\text{ heads})\prod_{k=1}^{n-1}P(C_k\text{ tails})\\ &=\frac{1}{2n+2}\prod_{k=1}^{n-1}\frac{2k+1}{2k+2}\\ &=\frac{1\cdot 3\cdot 5\cdots (2n-1)}{4\cdot 6\cdots (2n+2)} \end{align}$$

Since $P(X<\infty)=1$, we get: $\sum_{n=1}^{\infty} P(X=n) = 1$

More generally , give a sequence of real numbers, $a_k$ with $0\leq a_k\leq 1$ and $\sum_{k=1}^{\infty} a_k=+\infty$, then $\prod_{k=1}^{n} (1-a_k)\to 0$, and we get that $b_n=a_n\prod_{k=1}^{n-1}(1-a_k)$ satisfies $$\sum_{n=1}^{\infty} b_n = 1.$$

So if $a_{n}=\frac{1}{an+2}$ then $$b_n = \frac{1}{a+2}\frac{a+1}{2a+2}\cdots\frac{(n-1)a+1}{na+2}=\prod_{k=1}^{n} \frac{a(k-1)+1}{ak+2}$$

satisfies $\sum b_n = 1$.

More generally:

$$\sum_{n=1}^{\infty} \prod_{k=1}^{n} \frac{a(k-1)+b-1}{ak+b} = b-1$$

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