Here's something interesting I found out last year. I don't know if its been done before but I strongly suspect it has.

Consider a summation series of form \[S(n) = \sum_{k = 1}^{k = n}\frac{f(k)}{k!}\] such that \(f(1) = 1\).

Now let's do something unusual and create the following sequence starting with the term \(a_2 = 2\) and then \( a_{k+1} = (k + 1)(a_k + f(k))\) for all \(k > 2\).

Finally we'll focus our attention on the following product series \[P(n) = \prod_{k=2}^{k=n} ( 1 + \frac{f(k)}{a_k})\] for all \( n > 2\).

We'll show that \( S(n) = P(n) \) for all \(n > 2\).

**Proof : By induction on n.**
We'll start with the base case.
**Base Case: \(n = 2\)**

We have \(P(2) = 1 + \frac{f(2)}{a_2} = 1 + \frac{f(2)}{2} =\frac{f(1)}{1} + \frac{f(2)}{2} = S(2) \) as required.

**Induction Step**

This part is a quite long, so we'll break it down into 3 parts.

**Part 1 : Re-arranging the Factors of the Product Series**

We can write each factor from our product series as \[\begin{align} 1 + \frac{f(k)}{a_k} &= \frac{a_k}{a_k} + \frac{f(k)}{a_k} \\&= \frac{a_k + f(k)}{a_k} \\&= \frac{a_{k+1}}{(k+1)a_k} \end{align}\] by using the definition of our sequence.

**Part 2 : Re-writing our Product Series**

Now we have that \[ \begin{align} \prod_{k = 2}^{k = n} (1 + \frac{f(k)}{a_k}) &= \frac{a_{3}}{(3)a_2} \times \frac{a_{4}}{(4)a_3} \times ...\times \frac{a_{n+1}}{(n+1)a_n} \\ &= \frac{a_{n+1}}{a_2 \times 3 \times 4 \times ... \times n \times (n+1)} \\&=\frac{a_{n+1}}{(n+1)!} \end{align} \]

**Part 3 : Final Step**

Suppose \(P(q) = S(q) \) for some natural number \(q\).

Then\[ \begin{align} P(q+1) &= P(q)(1 + \frac{f(q+1)}{a_{q+1}}) \\&= P(q) + \frac{f(q+1)}{a_{q+1}}\times\frac{a_{q+1}}{(q+1)!} \\&= P(q) + \frac{f(q+1)}{(q+1)!} \\&= T(q) + \frac{f(q+1)}{(q+1)!} = T(q+1) \end{align} \] as required.

**End of Proof**

We'll go through some of the consequences of this in the next note!

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