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?

  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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


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: $$ \int_{0}^{\pi/2} \log\cos\theta\,d\theta = -\frac{\pi\log 2}{2}.$$

  • 1
    $\begingroup$ [+1] I applaude. $\endgroup$
    – Jean Marie
    Oct 8, 2017 at 17:08
  • 1
    $\begingroup$ Very insightful (+1). Can you please explain how you spot patterns like these and rearrange them? $\endgroup$ Oct 8, 2017 at 17:27
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ @hypergeometric Also, Jack does this every day all the time, and I guess practice makes you better $\endgroup$ Oct 8, 2017 at 17:43
  • 1
    $\begingroup$ Absolutely incredible, thank you:)) $\endgroup$ Oct 8, 2017 at 17:54

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 $A_{r+1}$.

  • $\begingroup$ 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

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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.