# Infinte series sum(n*(1/2)^n,n,k,infty)?

1. Dec 3, 2005

### cliowa

How can I calculate the value of this series as a function of k?

$$\sum_{n=k}^{\infty} n\cdot \big(\frac{1}{2}\big)^{n}$$

Plugging in numbers and trying to see a pattern doesn't work for me, I can't see it. I haven't found a way to decompose this sum into several ones where I can see what's going on.
Could anybody help me? It would be appreciated very much.

Best regards...Cliowa

2. Dec 3, 2005

### matt grime

you can sum x^n, so how about differentiating and summing nx^{n-1}?

we;ll ignore convergence issues and pretend that's ok. (please don't point out that you have n(1/2)^{n} and not n(1/2)^{n-1})

3. Dec 3, 2005

### cliowa

Yeah, maybe that'd work, but I really have to do it without differentiating!

4. Dec 3, 2005

### amcavoy

Why do you have to do it without? If you do it like matt grime says you get $\sum_{n=1}^{\infty}nx^{n}=\frac{x}{\left(1-x\right)^{2}}$.

5. Dec 3, 2005

### siddharth

As Matt and amcavoy pointed out, writing out the power series and then differentiaing it is an easy way.

Otherwise, you could treat this as an Arithmetic-Geometric Series.

So if

$$S = k (\frac{1}{2})^k + (k+1)(\frac{1}{2})^{k+1} + ....$$

$$\frac{S}{2} = k(\frac{1}{2})^{k+1} + (k+1)(\frac{1}{2})^{k+2} + ...$$

Subtract the two series and you will get a GP.

Last edited: Dec 4, 2005
6. Dec 4, 2005

### cliowa

Thanks alot. Could you be a bit more specific? What is a GP? If I subtract those to series I will get S/2, but that seems to be the same basic thing, with the indexes delayed. So what do I do with it?

7. Dec 4, 2005

### siddharth

No, it won't be the same thing.
You will have
$$S-\frac{S}{2} = k(\frac{1}{2})^k + (k+1-k)(\frac{1}{2})^{k+1} + (k+2-k-1)(\frac{1}{2})^{k+2} + ..$$
Write this to n terms and let n tend to infinity.
The series you now have is a Geometric Series.

8. Dec 4, 2005

### cliowa

Great! Thank you very, very much.
Simply out of interest: Could you also explain in a little more detail how I would do it with differentiating?

9. Dec 4, 2005

### matt grime

1+x+x^2+x^3+..

is the power series for (1-x)^-1

that is sufficient to solve it (ignoring convergence issues)

and if youi don't know what a GP is you're not going to get very far in doing these questions. Or was it you dfidn't knwo what the abbreviation was?

10. Dec 5, 2005

### cliowa

Thanks alot
Well, it was that I simply don't know the abbreviation! Would you mind revealing this secret to an outsider?
Best regards...Cliowa

11. Jul 9, 2008

### Take_it_Easy

An elementary approach.

Let's write the addendums of the serie one by one
$$k \left( \frac{1}{2} \right)^k$$
$$(k+1) \left( \frac{1}{2} \right)^{(k+1)}$$
$$(k+2) \left( \frac{1}{2} \right)^{(k+2)}$$
$$(k+3) \left( \frac{1}{2} \right)^{(k+3)}$$
$$(k+4) \left( \frac{1}{2} \right)^{(k+4)}$$
$$(k+5) \left( \frac{1}{2} \right)^{(k+5)}$$
and so on...
let's rewrite them in this "diagonal form" (notice that each row is the same of value of the rows wrote before)
$$k \left( \frac{1}{2} \right)^k$$
$$k \left( \frac{1}{2} \right)^{(k+1)} + \left( \frac{1}{2} \right)^{(k+1)}$$
$$k \left( \frac{1}{2} \right)^{(k+2)} + \left( \frac{1}{2} \right)^{(k+2)} + \left( \frac{1}{2} \right)^{(k+2)}$$
$$k \left( \frac{1}{2} \right)^{(k+3)} + \left( \frac{1}{2} \right)^{(k+3)} + \left( \frac{1}{2} \right)^{(k+3)} + \left( \frac{1}{2} \right)^{(k+3)}$$
$$k \left( \frac{1}{2} \right)^{(k+4)} + \left( \frac{1}{2} \right)^{(k+4)} + \left( \frac{1}{2} \right)^{(k+4)} + \left( \frac{1}{2} \right)^{(k+4)} + \left( \frac{1}{2} \right)^{(k+4)}$$
$$k \left( \frac{1}{2} \right)^{(k+5)} + \left( \frac{1}{2} \right)^{(k+5)} + \left( \frac{1}{2} \right)^{(k+5)} + \left( \frac{1}{2} \right)^{(k+5)} + \left( \frac{1}{2} \right)^{(k+5)} + \left( \frac{1}{2} \right)^{(k+5)}$$
and so on...
Now let's do the sum by columns! OK?
(In this moment there is a question that I will take back later)
What do we get when we do the sum by columns??? (see next post!)

12. Jul 9, 2008

### Take_it_Easy

We get this value from first column

$$k \sum_{i=k}^\infty \left( \frac{1}{2} \right)^i$$

this value from the second column

$$\sum_{i=k+1}^\infty \left( \frac{1}{2} \right)^i$$

this value from the third column

$$\sum_{i=k+2}^\infty \left( \frac{1}{2} \right)^i$$

this value from the fourth column

$$\sum_{i=k+3}^\infty \left( \frac{1}{2} \right)^i$$

this value from the fifth column

$$\sum_{i=k+4}^\infty \left( \frac{1}{2} \right)^i$$

and so on...

Notice that we from the $$(n+1)$$-th column you have exactly that value

$$\sum_{i=k+n}^\infty \left( \frac{1}{2} \right)^i$$

Now using the known formula (if you don't know just ask and I'll make it clear to you!)

$$\sum_{i=m}^{\infty}\left( \frac{1}{2}\right)^m = 2 \left(\frac{1}{2} \right)^m$$

we get that the value of the sum along the first column IS

$$2 k \left(\frac{1}{2} \right)^{k}$$

from second column we get the value

$$2 \left(\frac{1}{2} \right)^{(k+1)}$$

from the third

$$2 \left(\frac{1}{2} \right)^{(k+2)}$$

in short for the $$(n+1)$$-th column (for $$n >1$$) we get

$$2 \left(\frac{1}{2} \right)^{(k+n)}$$

we are CLOSE to conclusion (in next post)

13. Jul 9, 2008

### Take_it_Easy

Summing the values we obtained from 1st, 2nd, 3rd, 4th, 5th...and so on.. columns we have to do this infinite sum

$$k \left( \frac{1}{2} \right)^k + 2 \left( \frac{1}{2} \right)^{(k+1)} + 2 \left( \frac{1}{2} \right)^{(k+2)} + 2 \left( \frac{1}{2} \right)^{(k+3)} + 2 \left( \frac{1}{2} \right)^{(k+4)} + \cdots 2 \left( \frac{1}{2} \right)^{(k+n)} + \cdots =$$

$$= k \left( \frac{1}{2} \right)^k + 2 \left( \frac{1}{2} \right)^k \left[ \sum_{i=1}^{\infty} \left( \frac{1}{2} \right)^i \right] =$$

$$= k \left( \frac{1}{2} \right)^k + 2 \left( \frac{1}{2} \right)^k \left( \frac{1}{2} \right) = k \left( \frac{1}{2} \right)^k + \left( \frac{1}{2} \right)^k =$$
$$= (k+1) \left( \frac{1}{2} \right)^k$$

And that's it!

(I hope I haven't made some error but the method works!)

And now remember the notice I told I would take back later? Now it the moment.
To calculate the sum we decided to sum the addenda in a different way.
YOU can do such a passage ONLY if you know A PRIORI that the serie converges!
Commutativity for infinite (denumerable) elements works only on convergent series!
(By the way it is easy to see that this serie converges!)

It is just for Math's sake I wanted to specify this to you!

Hope you liked my effort to find the sum without derivating some serie term by term!
:)

14. Jul 10, 2008

### maze

You could also take the result for the geometric series, insert a parameter in the exponent, and then differentiate by that parameter to achieve the result. (alpha smaller than 1 for convergence (1/2 here))
$$\sum_{n=0}^{\infty} \alpha^n = \frac{1}{1-\alpha}$$

$$\sum_{n=0}^{\infty} \alpha^{t n} = \frac{1}{1-\alpha^t}$$

$$\sum_{n=0}^{\infty} \frac{d}{dt}\alpha^{t n} = \frac{d}{dt}\frac{1}{1-\alpha^t}$$

$$\sum_{n=0}^{\infty} n \alpha^{n} = \frac{\alpha}{(1-\alpha)^2}$$

For the sum starting at k instead of 0, do the same thing for the geometric series from 0 to k-1, then subtract.

Edit: oh, looks like matt already said this.