Solving infinite series

dublet · 2005-12-16 03:43:58

Given the following:

How would you solve the following summation?

mathsyperson · 2005-12-16 05:40:21

I plotted the first few hundred points of a_n, and it seems that the function diverges away from 0 in the negative direction. So, if you were to sum the infinite series, you'd just get - ∞.

I think you might have made a mistake with the function, though. As it is, it involves dividing by 0 when n = 1, so that indicates that something might be wrong. Plus, -∞ isn't a very satisfying answer.

dublet · 2005-12-16 07:55:10

mathsyperson wrote:

I plotted the first few hundred points of a_n, and it seems that the function diverges away from 0 in the negative direction. So, if you were to sum the infinite series, you'd just get - ∞.

The

, as you suggested, indeed.

I think you might have made a mistake with the function, though. As it is, it involves dividing by 0 when n = 1, so that indicates that something might be wrong. Plus, -∞ isn't a very satisfying answer.

This is no mistake, it's exactly what I got for one of my math exams.

The thing which I want to know is how you solve a summation, as all the books I have don't really show me a structured way to do it.

Another summation

Do you just plot a few points to see where it goes, and then you see if it goes to infinity, 1, 0, -1 or -infinity, or is there a more precise method (which is easy to understand )?

mathsyperson · 2005-12-16 08:37:31

Not that I know of. Your second example will have a more interesting answer, though, because it converges to 0, meaning that its sum will be finite.

[Disclaimer: I'm not at all good with LaTeX, so some of this stuff is likely to come out badly.]

There is a standard formula that we can use to help us:

We can split the summation up and rearrange it so that we can use this formula on it:

However, because the starting value is 3, and the above formula works with a starting value of 1, we need to take away the summation from n=1 to n=2 to compensate.

Using the above formula, this works out to be:

7/0.5 - 1/0.8 - (0.75*7/0.5 - 0.96/0.8) = 3.45

And there you go. The summation of the infinite series is 3.45

I probably could have said that 5 times faster if I wasn't trying to use LaTeX.

dublet · 2005-12-16 09:08:28

mathsyperson wrote:

There is a standard formula that we can use to help us:

With an infinite series, wouldn't this amount to the following?

And as such, for example:

Or am I missing something?

mathsyperson · 2005-12-16 09:15:00

That's right, but as (1/5)^∞ = 0, that's the same as 1/(4/5) = 5/4.

dublet · 2005-12-16 09:38:02

Point taken.

azdude121 · 2005-12-16 09:56:40

I am completely lost...

mathsyperson · 2005-12-16 10:42:15

That's just because the maths discussed here is at A-level level and you're in year 7 and so haven't learnt this yet.

MathsIsFun · 2005-12-16 11:08:31

For anyone who may be a bit lost "∑" just means to add up a lot!

Example:

It says to use the numbers from 1 to 5 (the n=1 at the bottom tells where to start, and the 5 at the top tells where to end), and whatever you see after the "∑" is what to do with the numbers before adding them (in this case multiply by 2)

So it is just saying: "2×1 + 2×2 + 2×3 + 2×4 + 2×5"

krassi_holmz · 2005-12-17 00:18:42

Sorry, if I made a mistake, but I think:

But
, so

and

Thus we have:
S=7(1-1/2-1/4)-(1/4-1/5-1/25)=7/4 - 1/100 = 87/50 = 1,74

Last edited by krassi_holmz (2005-12-17 00:55:20)

mathsyperson · 2005-12-17 06:54:52

Your mistake is here:

The first

in the numerator shouldn't be there.

krassi_holmz · 2005-12-17 07:15:24

2+4+8= (8-1)/(2-1) or 2+4+8=2(8-1)/(2-1), e.a
14=7 or 14=14?

I'm sure that

Why first x in the numerator shouldn't be there?

Last edited by krassi_holmz (2005-12-17 07:17:13)

mathsyperson · 2005-12-17 07:30:16

The formula you're thinking of is:

In this case, n = 1, so the specific formula would be:

krassi_holmz · 2005-12-17 08:06:55

Why are we disputing something fundamental?

krassi_holmz · 2005-12-17 08:08:54

http://mathworld.wolfram.com/GeometricSeries.html
Are we clear now?

mathsyperson · 2005-12-17 08:21:30

Wow. Sorry. I checked the textbook that I was reading, and I misread the formula. You're absolutely right, which means that your answer of 1.74 is right too.

krassi_holmz · 2005-12-17 08:29:38

And what did everybody draw from this lesson?
was here anything else posted? I don't think so.

[Yes. Yes there was. *points to 'last edited by...' tag*]

Last edited by krassi_holmz (2005-12-17 09:07:31)

mathsyperson · 2005-12-17 08:38:28

Except that you made the mistake of saying that you don't make mistakes. That means that when you do make a mistake, you'll look silly because you said that you don't make mistakes, but you would have just made a mistake. That's a mistake, so now you look silly because you said that you don't make mistakes, but you just made a mistake, you mistake maker, you.

krassi_holmz · 2005-12-17 09:04:21

Me what?
What a mistake?

mathsyperson · 2005-12-17 09:13:03

The mistake of trying to make a mod look silly, of course.

MathsIsFun · 2005-12-17 09:21:59

So, we could replace the last 10 posts with mathsy saying "Yes, you were right" ?

krassi_holmz · 2005-12-17 09:26:11

Why did you moderated me so brutally?

And now earnest.
Yes, your answer is just great! You deserve to be a moderator!(this was about the mathsy)

Last edited by krassi_holmz (2005-12-17 09:30:03)

Clayton · 2008-02-14 13:53:03

You guys should really make it as simple as possible... It's just a geometric series....

(1 + (r)^2 + (r)^3 +. . . + (r)^n ) converges IF : |r| < 1

then the infinite series converges to 1/(1-r).

You can easily check this to find what SERIES: (7/(2^i) - (1/(5^i)) converges to. Although the series starts
at i=3, all you would need to do is pull out a 1/8 for the 1/(2^i) series and 1/125 for the 1/(5^i) series. I hope the rest is clear.;)

needssumreview · 2008-10-15 18:33:12

I think you'd actually have to pull out ((1/2)^3)*((1/2)^2)*(1/2) for the first to return the index to 0 and not 2 but as my name says I haven't done this in a while and am here looking up how it is done for my convolution sums homework. hehe ^^

Math Is Fun Forum

#1 2005-12-16 03:43:58

Solving infinite series

#2 2005-12-16 05:40:21

Re: Solving infinite series

#3 2005-12-16 07:55:10

Re: Solving infinite series

#4 2005-12-16 08:37:31

Re: Solving infinite series

#5 2005-12-16 09:08:28

Re: Solving infinite series

#6 2005-12-16 09:15:00

Re: Solving infinite series

#7 2005-12-16 09:38:02

Re: Solving infinite series

#8 2005-12-16 09:56:40

Re: Solving infinite series

#9 2005-12-16 10:42:15

Re: Solving infinite series

#10 2005-12-16 11:08:31

Re: Solving infinite series

#11 2005-12-17 00:18:42

Re: Solving infinite series

#12 2005-12-17 06:54:52

Re: Solving infinite series

#13 2005-12-17 07:15:24

Re: Solving infinite series

#14 2005-12-17 07:30:16

Re: Solving infinite series

#15 2005-12-17 08:06:55

Re: Solving infinite series

#16 2005-12-17 08:08:54

Re: Solving infinite series

#17 2005-12-17 08:21:30

Re: Solving infinite series

#18 2005-12-17 08:29:38

Re: Solving infinite series

#19 2005-12-17 08:38:28

Re: Solving infinite series

#20 2005-12-17 09:04:21

Re: Solving infinite series

#21 2005-12-17 09:13:03

Re: Solving infinite series

#22 2005-12-17 09:21:59

Re: Solving infinite series

#23 2005-12-17 09:26:11

Re: Solving infinite series

#24 2008-02-14 13:53:03

Re: Solving infinite series

#25 2008-10-15 18:33:12

Re: Solving infinite series

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