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What is the formula for following sequence?

$$\frac12 + \frac12 \cdot \frac34 + \frac12 \cdot \frac34 \cdot \frac56 + ... + \frac12 \cdot \frac34 \cdot \frac56 ... \frac{2n - 1}{2n}$$

This is a question from my Calculus class about sequences.

Mauro ALLEGRANZA
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TenzoNakami
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4 Answers4

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$$ \begin{align} &\, 1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1) = (2n-1)!! \,,\quad\, 2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n) = (2n)!! \\[2mm] &\, S_{\small n} = \frac{1}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[6mm] &\, \color{red}{1+S_{\small n}} = 1+\frac{1}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm] &\, \qquad = \frac{1\,\cdot\,3}{2}+\frac{1\,\cdot\,3}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm] &\, \qquad = \frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4}+\frac{1\,\cdot\,3\,\cdot\,5}{2\,\cdot\,4\,\cdot\,6}+\,\cdots\,+\frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n-1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} \\[3mm] &\, \qquad = \,\cdots\, = \frac{1\,\cdot\,3\,\cdot\,5\,\cdots\,(2n+1)}{2\,\cdot\,4\,\cdot\,6\,\cdots\,(2n)} = \color{red}{\frac{(2n+1)!!}{(2n)!!}} \,\Rightarrow\, S_{\small n}=\frac{(2n+1)!!}{(2n)!!}-1 \\[6mm] &\, 1+\lim_{n\rightarrow\infty}S_{\small n} = \lim_{n\rightarrow\infty}\frac{(2n+1)!!}{(2n)!!}\color{red}{\,\longrightarrow\,\infty} \qquad \{{\small\text{divergent}}\} \end{align} $$

Hazem Orabi
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You are dealing with $$ \sum_{k=1}^{n}\frac{(2k-1)!!}{(2k)!!} = \sum_{k=1}^{n}\frac{1}{4^k}\binom{2k}{k}.\tag{1}$$ It is well-known that $\frac{1}{4^k}\binom{2k}{k}\sim\frac{1}{\sqrt{\pi k}}$ as $k\to +\infty$, hence the associated series is divergent by the p-test. We may also notice that, by the extended binomial theorem, $$ \sum_{k\geq 1}\frac{x^k}{4^k}\binom{2k}{k} = \frac{1}{\sqrt{1-x}}-1 \tag{2}$$ hence the RHS of $(1)$ is the coefficient of $x^n$ in the Taylor series (at $x=0$) of $$ \frac{1}{1-x}\left(\frac{1}{\sqrt{1-x}}-1 \right) = \frac{1}{(1-x)^{3/2}}-\frac{1}{1-x}\tag{3} $$ and by the extended binomial theorem (again): $$ \sum_{k=1}^{n}\frac{1}{4^k}\binom{2k}{k} = \color{red}{\frac{2n+1}{4^n}\binom{2n}{n}-1}.\tag{4} $$ Obviously $(4)$ can be proved through creative telescoping or probabilistic arguments, too. Interesting question.

Jack D'Aurizio
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  • Nice solution (+1, earlier). It can be shown easily that $$\frac 1{4^k}\binom {2k}k=\frac {2k+1}{4^k}\binom {2k}k-\frac {2(k-1)+1}{4^{k-1}}\binom {2(k-1)}{k-1}$$ which when summed telescopes nicely to the result shown in $(4)$. (as this question has been marked duplicate, it is not possible to post a new answer to show this derivation in full) – Hypergeometricx Feb 08 '17 at 17:40
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    @hypergeometric: of course. In terms of generating functions, we are just stating that the derivative of $\frac{1}{\sqrt{1-x}}$ is given by half the product between $\frac{1}{1-x}$ and the function itself, essentially. There also is a nice probabilistic argument for showing such telescopic property. – Jack D'Aurizio Feb 08 '17 at 18:00
  • The second term of RHS (in the equation in my comment) reduces to $$\frac {2k}{4^k}\binom {2k}k$$. – Hypergeometricx Feb 09 '17 at 02:44
  • How can we use Extended Binomial Theorem to prove (4)? – Brevan Ellefsen May 07 '17 at 03:18
  • The EBT allows to easily compute the Taylor coefficients of both the LHS and the RHS of (3). – Jack D'Aurizio May 07 '17 at 16:32
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Hint

$$\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot ...\cdot \frac{2n-1}{2n}= \frac{(2n-1)!}{2^n n!}$$

Surb
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If $x=-1,n=-1/2$ then your series is $$nx+\frac{n(n-1)}{2!}x^{2}+\cdots$$ which is the general binomial series for $(1+x)^{n}$ except for the missing first term $1$. It is known from theory of binomial series (see towards the end of this blog post) that if $x=-1$ then the binomial series is convergent only when $n\geq 0$. Thus the given series diverges.

Paramanand Singh
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