Summation Series to Product Series Part 2.

In the first part of these notes we showed the following $S(n) = \sum_{k = 1}^{k = n}\frac{f(k)}{k!} = \prod_{k = 2}^{k = n} (1 + \frac{f(k)}{a_k}) = P(n)$ for all $n > 1$ where the sequence $(a_k)$ is given by the recurrence relation $a_2 = 2$ and $a_{k+1} = (k+1)(a_k + f(k))$.

In this note we'll have a look at some applications of this.

Part 1 : A couple of Examples

For this constant function we have $S(n) = \sum_{k = 1}^{k = n}\frac{1}{k!}$. So $\prod_{k = 2}^{k \rightarrow \infty} (1 + \frac{1}{a_k}) = e-1$ where $a_2 = 2$ and $a_{k+1} = (k+1)(a_k +1)$ by considering Taylor expansion of $e$.

Here is another example : we can write the harmonic series as $\sum_{n=1}^{n \rightarrow \infty} \frac{1}{n}$. So $\prod_{k = 2}^{k \rightarrow \infty} (1 + \frac{(k-1)!}{a_k})$ where $a_2 = 2$ and $a_{k+1} = (k+1)(a_k +(k-1)!)$

Part 2 : Questions and Remarks for Brilliant Here's some questions I have for the Brilliant Community.

1) We showed that we can re-write the harmonic series as $\prod_{k = 2}^{k \rightarrow \infty} (1 + \frac{(k-1)!}{a_k})$. Can we prove that this diverges without assuming the summation series $\sum_{n=1}^{n \rightarrow \infty} \frac{1}{n}$?

2) Can we re-write the sequence $a_2=2$ and $a_{k+1} = (k+1)(a_k +f(k))$ with out recursion? Do these recursive sequences have any combinotorical properties?

3) We haven't really learnt anything significantly new from re-writing these series and in product series form this is computationally difficult due to the recursive sequences.

4) Post any interesting examples you can re-write below!

Note by Roberto Nicolaides
5 years, 11 months ago

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