# 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:

$${s}_{n}=\sum _{{i}_{0}}^{n-1}\frac{1}{{a}_{i}}$$ |

We would like to show that ${lim}_{n\to \mathrm{\infty}}{s}_{n}=1$. Putting over a common denominator, we obtain

$$ |

Define ${b}_{n}$ as follows:

$$ |

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

$${s}_{n}=\frac{{b}_{n}+1}{{a}_{n}-1}$$ |

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

${b}_{n+1}-1$ | $=$ | $$ | ||

$=$ | $({a}_{n}-1)+{a}_{n}({b}_{n}-1)$ |

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^{}:

${b}_{n}$ | $=$ | ${b}_{2}{\displaystyle \prod _{i=2}^{n-1}}{a}_{i}$ | ||

$=$ | $\frac{{b}_{2}}{{a}_{0}{a}_{1}}}{\displaystyle \prod _{1=0}^{n-1}}{a}_{i$ | |||

$=$ | $\prod _{1=0}^{n-1}}{a}_{i$ | |||

$=$ | ${a}_{n}-1$ |

Substituting this in the expression for ${s}_{n}$ yields

$${s}_{n}=\frac{{a}_{n}}{{a}_{n}-1}.$$ |

Since ${lim}_{n\to \mathrm{\infty}}{a}_{n}=\mathrm{\infty}$, it follows that ${lim}_{n\to \mathrm{\infty}}{s}_{n}=1$.

Title | sum of reciprocals of Sylvester’s sequence |
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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 |