Solving a Sequence to Find an Equation

  • Thread starter Thread starter Phyzwizz
  • Start date Start date
  • Tags Tags
    Sequence
AI Thread Summary
The discussion revolves around finding an equation to represent a specific sequence: 1, -1/2, 1/6, -1/24, 1/120. Initially, the user struggles to derive the first term using a recursive method based on multiplying the previous term by 1/n. Clarification is provided that the sequence is actually part of an infinite series, prompting a focus on finding the nth term. The breakthrough comes with the realization that the terms can be expressed using factorials, leading to the equation an = 1/n! * (-1)n+1. The user expresses frustration at not recognizing the solution sooner.
Phyzwizz
Messages
58
Reaction score
0
I know this problem probably extremely easy, and I can see the pattern of the sequence but I can't seem to figure out how to represent that pattern as an equation.

The sequence is 1-1/2+1/6-1/24+1/120...
Now I obviously see that to get the next term you multiply the previous term by 1/n
but doing this doesn't allow me to obtain "1" the very first term because I don't know the terms before it making all terms after it uncertain through this method.
The closest equation I've gotten is :

an=(1/(n(an-1)))(-1)n+1
using this I can obtain everything but the first term 1. I'm certain there must be another equation that is just not clicking in my head.
 
Physics news on Phys.org
Phyzwizz said:
I know this problem probably extremely easy, and I can see the pattern of the sequence but I can't seem to figure out how to represent that pattern as an equation.

The sequence is 1-1/2+1/6-1/24+1/120...
That isn't a sequence - it's an infinite series. A sequence is a list of numbers. In this case the sequence is {1, -1/2, 1/6, -1/24, 1/120, ...}

An infinite series is the sum of the terms in a sequence.

What are you trying to do in this problem, find the nth term in the sequence or add up the terms in the series? I would imagine that all you need to do is find the equation of the nth term in the sequence.
Phyzwizz said:
Now I obviously see that to get the next term you multiply the previous term by 1/n
but doing this doesn't allow me to obtain "1" the very first term because I don't know the terms before it making all terms after it uncertain through this method.
The closest equation I've gotten is :

an=(1/(n(an-1)))(-1)n+1
using this I can obtain everything but the first term 1. I'm certain there must be another equation that is just not clicking in my head.

Do you know about the factorial function? n! = 1 * 2 * 3 * ... * (n - 1) * n.
 
For this problem I am supposed to put the infinite series in sigma notation.
Yes I am aware of factorials and wow I just solved it. I can't believe I missed that, that's terrible.

an=1/n!(-1)n+1...

Wow I'm really angry at myself.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

Similar threads

Back
Top