sum of reciprocals of Sylvester’s sequence

We will show that the sum of the reciprocals of the Sylvester numbers indeed converges to 1.

Let $s_{n}$ denote a partial sum of the series of reciprocals:

s_{n}=\sum_{i_{0}}^{n-1}{1\over a_{i}}

We would like to show that $\lim_{n\to\infty}s_{n}=1$ . Putting over a common denominator, we obtain

s_{n}={\sum_{j=0}^{n-1}\prod\limits_{i\neq j\atop 0\leq i<n}a_{i}\over\prod_{i% =0}^{n-1}a_{i}}.

Define $b_{n}$ as follows:

b_{n}=1+\sum_{j=0}^{n-1}\prod_{i\neq j\atop 0\leq i<n}a_{i}

Using this new definition and the definition of the Sylvester numbers, we can rewrite the expression for $s_{n}$ as follows:

s_{n}={b_{n}+1\over a_{n}-1}

Let us now consider this sequence $b_{n}$ . We will start by deriving a recurrence relation:

	$\displaystyle b_{n+1}-1$	$\displaystyle=$	$\displaystyle\sum_{j=0}^{n}\prod_{i\neq j\atop 0\leq i<n+1}a_{i}=\prod_{i=0}^{% n-1}a_{i}+a_{n}\sum_{j=0}^{n-1}\prod_{i\neq j\atop 0\leq i<n}a_{i}$
		$\displaystyle=$	$\displaystyle(a_{n}-1)+a_{n}(b_{n}-1)$

Simplifying, we have $b_{n+1}=a_{n}b_{n}$ . Now, $b_{2}=1+a_{0}+a_{1}=6$ , hence we can solve the recursion with a product:

$\displaystyle b_{n}$	$\displaystyle=$	$\displaystyle b_{2}\prod_{i=2}^{n-1}a_{i}$
	$\displaystyle=$	$\displaystyle{b_{2}\over a_{0}a_{1}}\prod_{1=0}^{n-1}a_{i}$
	$\displaystyle=$	$\displaystyle\prod_{1=0}^{n-1}a_{i}$
	$\displaystyle=$	$\displaystyle a_{n}-1$

Substituting this in the expression for $s_{n}$ yields

s_{n}={a_{n}\over a_{n}-1}.

Since $\lim_{n\to\infty}a_{n}=\infty$ , it follows that $\lim_{n\to\infty}s_{n}=1$ .

Title	sum of reciprocals of Sylvester’s sequence
Canonical name	SumOfReciprocalsOfSylvestersSequence
Date of creation	2013-03-22 15:48:33
Last modified on	2013-03-22 15:48:33
Owner	rspuzio (6075)
Last modified by	rspuzio (6075)
Numerical id	7
Author	rspuzio (6075)
Entry type	Proof
Classification	msc 11A55