Summation Series to Product Series Part 2.

In the first part of these notes we showed the following S(n)=k=1k=nf(k)k!=k=2k=n(1+f(k)ak)=P(n) 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>1n > 1 where the sequence (ak)(a_k) is given by the recurrence relation a2=2 a_2 = 2 and ak+1=(k+1)(ak+f(k))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)=k=1k=n1k!S(n) = \sum_{k = 1}^{k = n}\frac{1}{k!} . So k=2k(1+1ak)=e1 \prod_{k = 2}^{k \rightarrow \infty} (1 + \frac{1}{a_k}) = e-1 where a2=2 a_2 = 2 and ak+1=(k+1)(ak+1)a_{k+1} = (k+1)(a_k +1) by considering Taylor expansion of ee.

Here is another example : we can write the harmonic series as n=1n1n\sum_{n=1}^{n \rightarrow \infty} \frac{1}{n} . So k=2k(1+(k1)!ak) \prod_{k = 2}^{k \rightarrow \infty} (1 + \frac{(k-1)!}{a_k}) where a2=2 a_2 = 2 and ak+1=(k+1)(ak+(k1)!)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 k=2k(1+(k1)!ak) \prod_{k = 2}^{k \rightarrow \infty} (1 + \frac{(k-1)!}{a_k}) . Can we prove that this diverges without assuming the summation series n=1n1n\sum_{n=1}^{n \rightarrow \infty} \frac{1}{n} ?

2) Can we re-write the sequence a2=2a_2=2 and ak+1=(k+1)(ak+f(k))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
4 years, 8 months ago

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