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Infinite Summation

  • Thread starter Peter G.
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  • #1
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I posted this initially in Pre-Calculus, but, seeing several people with similar doubts have posted this question here, I'll do the same, even though I haven't started Calculus

I am investigating the sum of infinite sequences.

such that: t0 = 1 and tn = (x ln (a))n/n!

They tell me to first consider the following sequences of terms:

1, (ln 2)/1, (ln 2)2/2x1, (ln 2)3/3x2x1

They then ask me to calculate the sum Sn of the first n terms of the sequence for when 0 is bigger or equal to 0 and smaller or equal to 10.

I couldn't however, find any relationship between the terms that indicates whether the sequence is arithmetic or geometric.

So are they asking me to simply grab the calculator, calculate the values and write them down?

How do I sum?

Thanks,
Peter G.
 

Answers and Replies

  • #2
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Which of the following do you mean:

[tex]t_n=\frac{ln(a)n}{n!}~\text{or}~t_n=\frac{(ln(a))^n}{n!}[/tex]

The second one would seem more logical, since it has a nicer answer.

The sequence is neither geometrical nor arithematical. You CAN find the sum, but only with some knowledge of calculus...

What they want you to do is grab a calculator, calculate the sum of the first 10 terms and see if you get something nice (you WILL get something nice!! at least for the second one).

So try the sum of [tex]\frac{(ln(a))^n}{n!}[/tex] for the first 10 terms for a=2, a=3, a=10. Do you get something nice?
 
  • #3
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Hi micromass, thanks for replying! :smile:

Sorry, I did in fact make a stupid formatting mistake. :redface: Yes, I am supposed to work with the second one.

I'll have to graph for n and then Sn.

I will try and then tell you how I went.

Thanks again,
Peter G.
 
  • #4
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This is the first time I do this in my calculator. And I got 2 as an answer for the sum. Is that correct?

If not, please tell me that I will keep trying.
 
  • #5
Dick
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When you do calculus, you learn how to show the sum of x^n/n! is the taylor series for e^x. So if you put x=ln(2) you get that the sum is e^(ln(2)). Does that confirm your calculator answer of 2?
 
  • #6
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I think so yes, but the sheet I have tells me to give my answers to 6 decimal places. I think that the values keep getting smaller and smaller and smaller, hence, closer and closer to 2 but never actually get there but my calculator simply rounds it.
 
  • #7
Dick
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I think so yes, but the sheet I have tells me to give my answers to 6 decimal places. I think that the values keep getting smaller and smaller and smaller, hence, closer and closer to 2 but never actually get there but my calculator simply rounds it.
That would be correct.
 
  • #8
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Indeed, the sum of the first 10 terms never really reaches 2. But your calculator just rounds it. But if you take an infinite number of terms, then you DO get exactly 2!!

Now, what happens with a=10??
 
  • #9
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When a = 10, with six decimal places I get 9.999702. (I am using excel now, easier than with calculator)

I am doing this exercise here: http://sspps.govoffice.com/vertical/Sites/%7B2B561619-723C-4585-8257-BD5E89887E15%7D/uploads/%7B8D4A401A-7DEA-4005-8BE3-4BC56AC913A0%7D.PDF [Broken]

On page 4.

It guides me step by step, and the first one is letter x = 1 and a = 2,
and then asks me to calculate the sum, give the answers to 6 decimal places and graph Sn, which, I am pretty sure I don't know how to do, but I will leave that for later.

After the steps it asks us for a suggestion of what happens to Sn as n approaches infinity
 
Last edited by a moderator:
  • #10
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Plotting is easy. You can even use excel for it.
Put n in one cell, and put [tex]S_n[/tex] in another cell, and then ask excel to plot the thing. You should see the [tex]S_n[/tex] approaching a certain value. For a=2, you have already shown it approaching 2. So that's what the graph should show...

Now, if n=10, then [tex]S_n[/tex] will be very close to 2. So what will happen if n approaches infinity?? It will come even closer to 2. So, we can say that if we take all the terms (=infinitely many), then the sum will exactly equal 2!!

Now, try the exercise for x=1 and a=3...
 
  • #11
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Thanks a lot guys, you are great :smile:

And, sorry for posting this on the Calculus section I was a bit confused, plus, I am a bit sick, got a headache and stuff... :frown:
 
  • #12
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Hi again guys,

I've now reached the less part of the first page. If you click the link with the exercise, you can see that they now ask us to analyse the general sequence where x = 1 and evaluate it for different values of a.

In this part, is it basically trying to prove that, no matter what value we choose for a, we will always only reach the value of a if we use all terms, infinitely many?

Thanks,
Peter G.
 
  • #13
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Hi again guys,

I've now reached the less part of the first page. If you click the link with the exercise, you can see that they now ask us to analyse the general sequence where x = 1 and evaluate it for different values of a.

In this part, is it basically trying to prove that, no matter what value we choose for a, we will always only reach the value of a if we use all terms, infinitely many?

Thanks,
Peter G.
I can't see anywhere where they ask that. It's also not true. For a=1, we have that [tex]\frac{\log(a)^n}{n!}=0[/tex] for all n. So already after the first term of the sum we get the final answer. It probably is true for [tex]a\neq 1[/tex]... But can you show me the exact question, since I cannot seem to find it in your pdf-file...
 
  • #14
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Yeah, you're right, it wouldn't work for 1. Well, I might have misinterpreted the question. I attached the part I am referring to:
 

Attachments

  • #15
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Yes, what they want you to do is

1) Do the same thing as you did before with some other values of a. Take a=1, a=10, a=345 and maybe some other values.

2) Finally they want you to draw a conclusion for the general series. If you take a an arbitrary number, then what will be the sum of [tex]\frac{(ln(a))^n}{n!}[/tex]?

I don't think they expect you to prove anything...
 
  • #16
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Ah ok, that's what I thought. What I did was I changed the value for a in my spreadsheet and, as you correctly, stated, as along as is not 1 or negative, the sum will always approach the value of a in a way it will only do so, when all the terms are used. So it is something else?
 
  • #17
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Ah ok, that's what I thought. What I did was I changed the value for a in my spreadsheet and, as you correctly, stated, as along as is not 1 or negative, the sum will always approach the value of a in a way it will only do so, when all the terms are used.
You'll have to be careful with the words you use. The sum will always approach the value of a, no matter how many terms are in the sum. However, the sum will only really equal a if all the terms are used. This is probably what you meant, but when dealing with series (= infinite sums), you need to be very careful.
 
  • #18
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Yeah, I always have a problem with the words in maths :tongue2: But yeah that's what I meant.

Thanks your for your patience, I think it is enough for me for today. :zzz:
 
  • #19
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Hi!

I just finished my investigation :smile:

So, for the last part, this is what I noticed: (I'll try and be accurate with words)

When changing the values of x for the same value of a, as Sn approaches infinity, it perpetually moves towards ax, only equalling it though, if all the terms are used, that, is an infinite amount.

So, for example. When a = 4 and x = 2. As n approaches infinity, Sn will increasingly approach 16.

They ask however, two things which I am sure of how to answer:

"Discuss the scope and/or limitations of the general statement:"

So, I believe what they want us to say is that this investigation only works for positive values of a, since we can't ln negative numbers because e is a positive base and, for when a or x are = 0 or when a = 1.

And:"Explain how you arrived at the general statement"

Well, apart from spending hours graphing what your paper told me :tongue:

I'm unsure here. I think they want us to reason our conclusion. For the second part I believe it should go more or less like this: (x ln a), according to one of the log laws, is = ln ax, which is what we are approaching...

Thanks,
Peter G.
 
  • #20
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I think that are very good answers!! Great job!
 
  • #21
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Cool thanks a lot for your help, you were patient and didactic throughout. :smile:
 
  • #22
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Hi again,

Sorry to bring this back after such a long time :tongue: but I have a simple doubt.

In the second page of the investigation, page 5 if I am not mistaken, tells us to test T9 (2,x) for example, for various values of x. They then ask us to plot the relation between T9 and x.

Carried away by the first part that wanted the relation between n and Sn I did the same for the second part, well, Sum up to T9 as n increased. But, this time, they want us to plot something different? Like, for example, if we use a = 2 as indicated and test for x = 2, 3, 4 and 5, we should have a graph that shows points: (2,4) (3,8) (4,16) (5,32)

Do you understand what I mean?

Thanks,
Peter G.
 
  • #23
22,097
3,278
Hi again,

Sorry to bring this back after such a long time :tongue: but I have a simple doubt.

In the second page of the investigation, page 5 if I am not mistaken, tells us to test T9 (2,x) for example, for various values of x. They then ask us to plot the relation between T9 and x.

Carried away by the first part that wanted the relation between n and Sn I did the same for the second part, well, Sum up to T9 as n increased. But, this time, they want us to plot something different? Like, for example, if we use a = 2 as indicated and test for x = 2, 3, 4 and 5, we should have a graph that shows points: (2,4) (3,8) (4,16) (5,32)

Do you understand what I mean?

Thanks,
Peter G.
Hmmm, I'm quite lost as why you have the points (2,4), (3,8),... You will want the points (2,T9(2)) and so on. So you'll have to calculate T9(2). Of course, infinitely many summations will give you 4, but the problem only asks for 9 summations!!
 
  • #24
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What I meant was:
Let's take a to be 2 and test for various values of x: x = 2, x =3, x = 4, x = 5
When x =2 and a = 2, the sum of the first 9 terms will move towards 4.
When x =2 and a = 3, the sum will move towards 8
for when a = 4, towards 16
for when a = 5, towards 32 and so on
If we plot the value of x against the value of the sum of the first 9 terms we will get points that are:

x = 2, y = 4
x = 3, y = 8
x = 4, y = 16
x = 5, y = 32

It will look like a exponent graph.

In the first part I graphed the position n, all the way from 0 to 10 against Sn:
For example, when a = 2 and x = 1. I had my first point in the graph:
x axis: 0, y axis: 1
x = 1, y = 1.693147
x = 2, y = 1.999374
and etc...

From your explanation, I guess I should do the graph like the one just above, like I did for the first part?
 
  • #25
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Yes, they don't want exact answers here. They don't want to know what it moves towards, they simply want to know the sum of the first 9 parts! So just make the sum and graph it, don't graph the exact answer (=the infinite summation), because this is not what they ask here...
 

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