Sum to infinity of a nongeometric series? impossible?

TheSetMiner

I need to calculate a moment generating function for a probability function and to do so I need to get the sum of x to infinity of the following expression:

((x-1)/4)*((2e^t/3))^x

for x = 2, 3, 4....

Hope that is clear.

Clearly, without the ((x-1)/4) term this would resolve to a fairly simple sum to infinity of geometric series calculation where a = ((2e^t/3))^2 & r=((2e^t/3)) but including the term complicates things I think because there is no longer a common r term to be found.
Anyone have any idea how to approach this?
Thanks

dlouth15

I put it into a computer algebra system and got

[latex]\displayvalue{\frac{e^{2 t}}{\left(2 e^t-3\right)^2}}[/latex]

as the sum from 2 to infinity.

It gives the sum from 2 to n as

[latex]\displaystyle{\frac{3^{-n} e^t \left(2^n \left(2 (n-1) e^t-3 n\right) \left(e^t\right)^n+2\ 3^n e^t\right)}{2 \left(3-2 e^t\right)^2}}[/latex]

I don't know how you would arrive at that but you might be able to prove by induction that it is true.

TheSetMiner

dlouth15 wrote: »

I put it into a computer algebra system and got

[latex]\displayvalue{\frac{e^{2 t}}{\left(2 e^t-3\right)^2}}[/latex]

as the sum from 2 to infinity.

It gives the sum from 2 to n as

[latex]\displaystyle{\frac{3^{-n} e^t \left(2^n \left(2 (n-1) e^t-3 n\right) \left(e^t\right)^n+2\ 3^n e^t\right)}{2 \left(3-2 e^t\right)^2}}[/latex]

I don't know how you would arrive at that but you might be able to prove by induction that it is true.

Thanks for your reply. So it is possible afterall. I assume there is some kind of formula to use but I can't seem to find it anywhere.


Can I ask what calculator you used? There may be a step-by-step option I can take a look at.

The answer you got is definitely correct as the question actually asks one to show how the sum can be expressed as:
[latex]\displayvalue{\frac{e^{2 t}}{\left(2 e^t-3\right)^2}}[/latex]

dlouth15

I used Mathematica. Unfortunately it doesn't break things down into steps.

If you wanted to show that [latex]\displayvalue{\frac{e^{2 t}}{\left(2 e^t-3\right)^2}}[/latex] is true, you could start by proving that

[latex]\displaystyle{\frac{3^{-n} e^t \left(2^n \left(2 (n-1) e^t-3 n\right) \left(e^t\right)^n+2\ 3^n e^t\right)}{2 \left(3-2 e^t\right)^2}}[/latex]

is true by induction.

i.e. show that it is true for n=2 - just put n=2 into the big expression and show that it is equal to ((x-1)/4)*((2e^t/3))^x for x=2.

Then show that if it is true for n, it is also true for n+1.

You have now proved the expression for the sum to n.

Then finally show that the limit as n goes to infinity is your final result.

It may not be the way they want to to do it, but it is mathematically valid.

TheSetMiner

dlouth15 wrote: »

I used Mathematica. Unfortunately it doesn't break things down into steps.

If you wanted to show that [latex]\displayvalue{\frac{e^{2 t}}{\left(2 e^t-3\right)^2}}[/latex] is true, you could start by proving that

[latex]\displaystyle{\frac{3^{-n} e^t \left(2^n \left(2 (n-1) e^t-3 n\right) \left(e^t\right)^n+2\ 3^n e^t\right)}{2 \left(3-2 e^t\right)^2}}[/latex]

is true by induction.

i.e. show that it is true for n=2 - this would just amount to showing that the first two terms added together equal that expression for n=2.

Then show that if it is true for n, it is also true for n+1.

You have now proved the expression for the sum to n.

Then finally show that the limit as n goes to infinity is your final result.

It may not be the way they want to to do it, but it is mathematically valid.

Thanks, looks like I will go about it this way as I have no other approach at the moment.
Just one question, isn't it a bit random to pluck that sum to n expression out of nowhere? or is it completely valid once proven?

dlouth15

TheSetMiner wrote: »

Thanks, looks like I will go about it this way as I have no other approach at the moment.
Just one question, isn't it a bit random to pluck that sum to n expression out of nowhere? or is it completely valid once proven?

It is strange to do it that way, but is valid once proven. At the end you will have shown the infinite sum expression to be true.

However I can't guarantee that it will be marked well.

Note I changed my last post since you replied.

TheSetMiner

dlouth15 wrote: »

It is strange to do it that way, but is valid once proven. At the end you will have shown the infinite sum expression to be true.

However I can't guarantee that it will be marked well.

Note I changed my last post since you replied.

Cheers. It's better than leaving it blank anyway. I'll do some googling and hopefully that formula pops up somewhere as a special geometric series sum to infinity expression or something.

Really appreciate the help, dlouth.

ZorbaTehZ

In your expression set n=x-1 to get
[latex]\frac C 4 nC^n[/latex]
where C=2exp(t)/3. For x<1, one has
[latex]\frac{1}{1-x}=\sum_{i=0}^{\infty}x^i[/latex]
Then differentiate both sides and multiply by x to obtain
[latex]\frac{x}{(1-x)^2}=\sum_{i=0}^{\infty}ix^i[/latex]
Then replace the values assuming t<ln3/2 and you get what was obtained my mathematica.

MathsManiac

It's an arithmetico-geometric sequence, as is more readily seen when it's in the form Zorba has it. There's a general technique for summing them to n, and hence to infinity when they converge.

http://en.wikipedia.org/wiki/Arithmetico-geometric_sequence