# 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

Define $b_{n}$ as follows:

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

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 SumOfReciprocalsOfSylvestersSequence 2013-03-22 15:48:33 2013-03-22 15:48:33 rspuzio (6075) rspuzio (6075) 7 rspuzio (6075) Proof msc 11A55