Infinite series, power series problem

[Liquidxlax]

Homework Statement

In a water purification process, one-nth of the impurity is removed in the first stage. In each succeeding stage, the amount of impurity removed is one-nth of that removed in the preceding stage. Show that if n=2, the water can be made a pure as you like, but if n=3, at least one-half of the impurity remains no matter how many stages are used.

Homework Equations

may be relevant, S_n = (a_o(1-rⁿ)/(1-r))S_n-rS_n = (a_o-a_or^n)

The Attempt at a Solution

well i figured out the series expansion for bothSUM (1/(n^2 -n)) (sorry i couldn't figure out the text thing)n=2 gives = 1/2 + 1/6 + 1/12 + 1/20 + ...

n=3 gives = 1/6 + 1/12 + 1/20 + 1/36 + ...

I know the sum of n=2 is pretty much 1 and n=3 is pretty much 1/2. My real problem is i can't figure out how to get it into:S_n = (a_o(1-rⁿ)/(1-r))so I can differentiate to get the sum. Any help would be appreciated and thanks for helping a noob

 

[jgens]

n=2 gives = 1/2 + 1/6 + 1/12 + 1/20 + ...

n=3 gives = 1/6 + 1/12 + 1/20 + 1/36 + ...

Can you explain how you got these sums? They don't model the situation that the problem describes.

 

[Liquidxlax]

Can you explain how you got these sums? They don't model the situation that the problem describes.

i got those sums, by finding the equation 1/(n-1) - (1/n) = 1/(n^2 - n)

This is what i did for n=2 n=2 wolfram

n=3 n=3 wolfram

 

[ehild]

Calculate the removed impurity.
At the first step, 1/n part of the impurity is removed.
In the second and every subsequent step, 1/n times the previously removed quantity is removed.

So the removed amount in k steps is

1/n +1/n^2 +1/n^3 +...1/n^k. What do you get after infinite steps?

ehild

 

[Liquidxlax]

Calculate the removed impurity.
At the first step, 1/n part of the impurity is removed.
In the second and every subsequent step, 1/n times the previously removed quantity is removed.

So the removed amount in k steps is

1/n +1/n^2 +1/n^3 +...1/n^k. What do you get after infinite steps?

ehild

i'm sorry but i messed up my second post, I'm not sure if you were responding to that one, but i made a mistake in my equation.

 

[jgens]

If you follow ehild's lead, you'll solve the problem pretty quickly. I'm not sure what reasoning your using to get your sums, but it's not correct.

 

[Liquidxlax]

can someone please just give me a solution, because this is apparently the easiest part of my assignment, and there are 3 questions like this, meanwhile i have completed all the apparent harder ones. If i could see one example of this sort, i should be able to figure it out.

 

[jgens]

It's against PF policy to give complete solutions. We can however, guide you towards a solution.

 

[Liquidxlax]

It's against PF policy to give complete solutions. We can however, guide you towards a solution.

okay that is something i did not know.

Well can you tell me why i am wrong, because I'm really lost

all i have to work with for equations are;

S_n = (a(1-rⁿ))/(1-r)

S = lim (n\rightarrow\infty) S_n

S = a/(1-r)

for this section, so this is what I'm assuming should help solve this problem.
I'm not sure how to find r or a, and i guess that is my major problem.

 

[jgens]

Use this ...

Calculate the removed impurity.
At the first step, 1/n part of the impurity is removed.
In the second and every subsequent step, 1/n times the previously removed quantity is removed.

So the removed amount in k steps is

1/n +1/n^2 +1/n^3 +...1/n^k. What do you get after infinite steps?

ehild

 

[Liquidxlax]

Use this ...

1/n^2 + 1/n^3 + 1/n^4 + ... + 1/n^(k-1) = S_n

1/n^3 + 1/n^4 + 1/n^5 + ... + 1/n^k-1 + 1/n^k = rS_n

S_n - rS_n = 1/n^3 - 1/n^k = a(1 - r^k)

a(1 - r^k) = (1/n^3)(1 - (1/n^k)) no k-3 since k is going to infinity?

S_n(1 - r) = S_n(1- (1/n))S_n = ((1/n^3)(1 - (1/n^k))) / (1 - (1/n))S = lim (k approaches infinity) S_n

S = (1/n^3)/(1 - (1/n))

Is this it? I think i got it

 

[jgens]

1/n^2 + 1/n^3 + 1/n^4 + ... + 1/n^(k-1) = S_n

What happened to the term 1/n?

 

[Liquidxlax]

What happened to the term 1/n?

whoops, but if i add that it should work?

i need n=2 to infinity to ~= 1

and n=3 to infinity to ~=1/2

 

[jgens]

You, if you redo your work including that term, it should work out.

 

[Liquidxlax]

You, if you redo your work including that term, it should work out.

okay screw it, i'll try it your way once more, and if i don't get it... then i'll just skip it

Thanks for trying to help me, I can't help I'm border line retarded sometimes

 

[ehild]

I think you mix the "n" in the problem with the "running index" of a sequence. Forget Sn. It is the sum of the first n term of a sequence. You need the sum of infinite terms which is S=a1/(1-r) for a geometric sequence with first term a1 and ratio r.

Consider the removed amount of the impurity. Suppose its original amount is 1 in some unit. In the first step, you remove 1/n of it. In the next step, you remove the n-th part of the previously removed amount, so you remove (1/n)^2. In the third step, you remove (1/n)^3, in the k-th step, the removed amount is (1/n)^k. You continue the process endlessly, and you get the following sequence

1/n, (1/n)^2, (1/n)^3... (1/n)^k...

If n=2, it is 1/2, 1/4, 1/8, 1/16...

For n=3: 1/3, 1/9, 1/27, 1/81...

What kind of sequence is it and what is the sum of infinite terms?

This sum is the theoretical amount of impurity you can remove in infinite steps.

ehild.

 

[Liquidxlax]

you know that is what i did exactly and right after my last post i realized that, and got the problem figured out. Thank you very much for the help and not getting mad because i was so stubborn