# Sum of (i^2)/(4^i) where i is from 0 to infinity.

1. Jun 21, 2004

### Johnny Leong

Sum of (i^2)/(4^i) where i is from 0 to infinity.

2. Jun 21, 2004

### NateTG

You mean $$\sum_{i=1}^\infty \frac{i^2}{4^i}$$? (I don't like $$\frac{0^2}{4^0}$$ - it's not necessarily clear what it is.)

Well, for $$i \geq 4$$ ,
$$2^i \geq i^2$$,
so
$$\frac{i^2}{4^i} < \frac{2^i}{4^i} = \frac{1}{2^i}$$
so
$$\sum_{i=1}^\infty \frac{i^2}{4^i} < \frac{1}{4}+\frac{4}{16}+\frac{9}{64} + \sum_{i=4}^{\infty} \frac{1}{2^i}$$

Perhaps you can find some better bounds?

3. Jun 21, 2004

### Muzza

I'm sorry, but what's unclear about 0^2/4^0?

4. Jun 21, 2004

### Gokul43201

Staff Emeritus
$$\sum_{i=1}^\infty \frac{i^2}{4^i} > \frac{1}{4}+\frac{4}{16}+\frac{9}{64} + \sum_{i=4}^{\infty} \frac{1}{4^i}$$

That gives a range of 0.7 to 0.9. Wonder if it isn't just 3/4 or 4/5 ?

5. Jun 21, 2004

Mathematica's saying 20/27 (~.740741). But, like most things Mathematica, I have no idea how it produced that number.

6. Jun 21, 2004

### Johnny Leong

Actually, this question could not give an accurate answer. The answer should just be an approximation, right? Because the terms in the summation are not having some sequence properties.

7. Jun 22, 2004

### Johnny Leong

I think Gokul43201's answer would not be a good one because after I have read a book, I found that when we do some approximation, the approximation's value had better dominate the value of the original question.

8. Jun 22, 2004

### HallsofIvy

Staff Emeritus
What in the world does that mean? What do you mean by "dominate the value of the original question? An approximation is an approximation. There exist good approximations and bad approximations. The ideal is to get as close as possible to the true value for the work done.

9. Jun 22, 2004

### arildno

$$F(x)=\sum_{i=0}^{\infty}i^{2}x^{i}$$
$$\sum_{i=0}^{\infty}i^{2}x^{i}=\sum_{i=0}^{\infty}xi\frac{d}{dx}x^{i}$$
$$\sum_{i=0}^{\infty}xi\frac{d}{dx}x^{i}=\frac{d}{dx}\sum_{i=0}^{\infty}ix^{i+1}-\sum_{i=0}^{\infty}ix^{i}$$
$$\sum_{i=0}^{\infty}ix^{i}=x\frac{d}{dx}\sum_{i=0}^{\infty}x^{i}$$
Hence, we have:
$$F(x)=\frac{d}{dx}(x^{2}\frac{d}{dx}\frac{1}{1-x})-x\frac{d}{dx}\frac{1}{1-x}$$
or:
$$F(x)=\frac{x^{2}+x}{(1-x)^{3}}$$
The sum of the original series is found by evaluating $$F(\frac{1}{4})$$

Last edited: Jun 22, 2004
10. Jun 22, 2004

### Gokul43201

Staff Emeritus
Which is 20/27, as Mathematica magically guessed.

Nice, arildno.

11. Jun 22, 2004

A very clever little trick, indeed.

12. Jun 23, 2004

### Johnny Leong

I mean the approximation should be an upper bound to the accurate answer to the original question.

13. Jun 23, 2004

### Johnny Leong

What are you doing, arildno? You are professional but I do not understand. Why you do like this?

14. Jun 23, 2004

### arildno

OK, I'll break up this in tiny pieces; then pin-point what you don't understand.

1. Changing perspective from number to function:
We start out with the series:
$$S=\sum_{i=0}^{\infty}i^{2}(\frac{1}{4})^{i}$$
Our aim is to find the number S!
However, this is difficult to do as it stands; what I want to do, is to change the problem slightly, so that:
a) I gain access to powerful solving techniques in the new problem to be solved
(which is not accessible in the original problem)
b) I can easily find the answer to the original problem once I have found the answer to the new problem

That is why I change perspectives to try to simplify the expression for the following function:
$$F(x)=\sum_{i=0}^{\infty}i^{2}x^{i}$$

a): Since I now have a power series function , I can use the extraordinarily powerful teqnique of differentiation to help me solve the new problem (simplifying the expression for F(x)!)
b) If I can do this, then I can find the answer to my original problem simply by
computing $$F(\frac{1}{4})$$

I'll let you ponder on this for a while, if you have some specific questions, pleasy notify..