Summation Series to Product Series Part 1.

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)=k=1k=nf(k)k!S(n) = \sum_{k = 1}^{k = n}\frac{f(k)}{k!} such that f(1)=1f(1) = 1.

Now let's do something unusual and create the following sequence starting with the term a2=2a_2 = 2 and then ak+1=(k+1)(ak+f(k)) a_{k+1} = (k + 1)(a_k + f(k)) for all k>2k > 2.

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

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

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

We have P(2)=1+f(2)a2=1+f(2)2=f(1)1+f(2)2=S(2)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 1+f(k)ak=akak+f(k)ak=ak+f(k)ak=ak+1(k+1)ak\begin{aligned} 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{aligned} by using the definition of our sequence.

Part 2 : Re-writing our Product Series

Now we have that k=2k=n(1+f(k)ak)=a3(3)a2×a4(4)a3×...×an+1(n+1)an=an+1a2×3×4×...×n×(n+1)=an+1(n+1)! \begin{aligned} \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{aligned}

Part 3 : Final Step

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

ThenP(q+1)=P(q)(1+f(q+1)aq+1)=P(q)+f(q+1)aq+1×aq+1(q+1)!=P(q)+f(q+1)(q+1)!=T(q)+f(q+1)(q+1)!=T(q+1) \begin{aligned} 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{aligned} as required.

End of Proof

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

Note by Roberto Nicolaides
4 years, 9 months ago

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