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=1∑k=nk!f(k) such that f(1)=1.
Now let's do something unusual and create the following sequence starting with the term a2=2 and then ak+1=(k+1)(ak+f(k)) for all k>2.
Finally we'll focus our attention on the following product series P(n)=k=2∏k=n(1+akf(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+a2f(2)=1+2f(2)=1f(1)+2f(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+akf(k)=akak+akf(k)=akak+f(k)=(k+1)akak+1 by using the definition of our sequence.
Part 2 : Re-writing our Product Series
Now we have that k=2∏k=n(1+akf(k))=(3)a2a3×(4)a3a4×...×(n+1)anan+1=a2×3×4×...×n×(n+1)an+1=(n+1)!an+1
Part 3 : Final Step
Suppose P(q)=S(q) for some natural number q.
ThenP(q+1)=P(q)(1+aq+1f(q+1))=P(q)+aq+1f(q+1)×(q+1)!aq+1=P(q)+(q+1)!f(q+1)=T(q)+(q+1)!f(q+1)=T(q+1) as required.
End of Proof
We'll go through some of the consequences of this in the next note!
#Series
#TaylorSeries
#Summation
#Productseries
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