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Infinite Summation

  • #1

Homework Statement



"Aim: In this task, you will investigate the sum of infinite sequences tn, where

tn = [tex]{\frac{(x\ln{a})^n}{n!}}[/tex], and t0=1

Consider the sequence when x=1 and a=2.

Using technology, plot the relation between Sn (the sum of t0+...+tn) and the first n terms of the sequence for [tex]{0}\leq{n}\leq{10}[/tex].

What does this suggest about Sn as n approaches [tex]\infty[/tex]?"


The Attempt at a Solution



We were given this assignment (which originally is much lengthier than this, but this is the core idea of it) with the prospect of using a calculator and/or graphing software to deduce that the value of the sum of this sequence converges as n becomes arbitrarily large. We have never been exposed to infinite sums (or even calculating the value of non-infinite sums) and are expected to do this completely dry, not using any formal analysis of the sum. I'm suspicious that this sum,

[tex]\sum\limits_{n=0}^{\infty}{\frac{(x\ln{a})^n}{n!}}[/tex]

has an analytic solution, and I'd love to find out what it is. I'm also taking AP Calculus BC as a parallel to this class, but we haven't come close to infinite sums yet. I've done a lot of integration and significantly more differential calculus independently. I was wondering, what direction should I take if I wanted to include an analytical solution for this convergence (not just plugging in values into a calculator and seeing where it leads, as my school wants me to)? Thanks, any help appreciated.
 

Answers and Replies

  • #2
hunt_mat
Homework Helper
1,720
16
let [tex]y=x\ln a=\ln (a^{x})[/tex], then you're talking about the series:
[tex]
\sum_{n=0}^{\infty}\frac{y^{n}}{n!}=e^{y}
[/tex]
 
  • #3
22,097
3,278
You are correct, there is an analytic solution: the sum of the series is [tex]ae^x[/tex].

So if x=1 and a=2, then the series simply converges to 2.

However, the proof of this fact entails Taylor series and other stuff. And if you have never seen series before, then I think that the proof is out of your grasp. I guess that the purpose of the assignment was that an infinite sum can give a finite answer and that this answer can be approximated...
 
  • #4
You are correct, there is an analytic solution: the sum of the series is [tex]ae^x[/tex].

So if x=1 and a=2, then the series simply converges to 2.

However, the proof of this fact entails Taylor series and other stuff. And if you have never seen series before, then I think that the proof is out of your grasp. I guess that the purpose of the assignment was that an infinite sum can give a finite answer and that this answer can be approximated...
Maybe I should have been more clear in my original post; I've seen sequences before but I've never done calculations that reduce the value of the sum to a concise little formula. I've seen the proofs for this being done but I've never learned any hard and fast methods of creating one of these short formulas, and I'd really like to. I'll still have to do the tedious plotting, but doing things ad-hoc like that drives me absolutely insane! My math teacher will surely get sick of me peppering him with questions.
 
  • #5
22,097
3,278
So you have seen sequences before, that's good. Have you seen the theorem of Taylor somewhere? Or Taylor series? Because that is what will help you are.

In general, giving an analytic answer to an infinite sum is impossible... It can only be done for very specific infinite sums. For example, nobody really knows what

[tex]\sum_{n=1}^{+\infty}{\frac{1}{n^3}}[/tex]

equals. And this is really an innocent looking sum. However, the sum that your teacher gave you is a very famous one. And you will certainly see it again. But you'll need some more calculus to show that the sum does in fact equal 2...
 
  • #6
22,097
3,278
  • #7
hunt_mat
Homework Helper
1,720
16
You are correct, there is an analytic solution: the sum of the series is [tex]ae^x[/tex].
If you read my post you would have realised it was a^{x} rather than ae^{x}
 
  • #8
22,097
3,278
If you read my post you would have realised it was a^{x} rather than ae^{x}
Yes, of course, stupid me :tongue:
 
  • #9
Thanks guys! I'm teaching myself Taylor polynomials and series now. I hope to have grasped the reasoning behind this summation before the end of the two-week session we have to work on this project, so I can put the actual analysis at the end of my report. I find it really pointless to give us such a task without having any sort of challenge; plugging numbers into a calculator for half a month doesn't sound like anything of that nature.

Edit: I understand it now, and can't wait to get to Taylor series in my Calculus class so I can formally learn it! Great stuff. Now, to destroy this terrible assignment..
 
Last edited:

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