sum of reciprocals of Sylvester’s sequence

We will show that the sum of the reciprocals of the Sylvester numbers indeedconverges 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 $\\lim_{n\\to\\infty}s_{n}=1$ . Puttingover a common denominator, we obtain

s_{n}={\\sum_{j=0}^{n-1}\\prod\\limits_{i\eq 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\eq 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 arecurrence relation:

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

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 $\\lim_{n\\to\\infty}s_{n}=1$ .