×

# 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
1 year, 8 months ago