# Converging geometric series

• Aristarchus_
It's a quadratic.-DanI am not sure what you did there. Could you please write your derivation in a more formal way?Do you know how to solve a quadratic equation?I am not sure what you did there. Could you please write your derivation in a more formal way?Do you know how to solve a quadratic equation?I can start with$$S = a_1 \cdot (k^{n} - 1/ k-1)$$This is not correct. You are missing two parentheses.Also, why did you set $n = 1$?This is not correct. You are missing two parentheses.Also, why did you set $n = 1$?f

#### Aristarchus_

Homework Statement
In a convergent geometric series with positive terms, the sum of the first and the third term is equal to the product of the first and the second term. (As this makes any sense...I can make the case where this is not true)
Find the exact values for the quotient ##k## in the series and the first term ##a_1## when the sum of the series is the smallest possible.
Relevant Equations
The solution is ##k = \sqrt{2}-1## and ##a_1 = 2 \cdot \sqrt{2}##
I do not have any reasonable attempts at this problem, as I am trying to figure out how one can get the correct answer when we are not given any values. Maybe if some of you sees a mistake here, that implies that the values from the previous example should be used...

##a_3 = a_1 \cdot k{2}## and ##a_2=a_1 \cdot k##
##a_1 + a_3 = a_1 \cdot (1+k^{2})## ##a_1 \cdot a_2 = a_1^{2} \cdot k##
from the problem, it then follows that ##a_1 \cdot (1+k^{2}) = a_1^{2} \cdot k##
But I do not see any connections yet...

What is meant by the smallest possible sum?

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You should fix the Latex. You need two hashes, not a single dollar.

The problem is tricky, perhaps. You must at least be able to write down the constraints you are given for what is a minimisation problem.

Relevant Equations:: The solution is k = ##\sqrt{2}-1## and ##a_1 = 2 \sqrt{2}##

Homework Statement:: The problem reads as follows
In a convergent geometric series with positive terms, the sum of the first and the third term is equal to the product of the first and the second term. (As this makes any sense...I can make the case where this is not true)
Find the exact values for the quotient $k$ in the series and the first term $a_1$ when the sum of the series is the smallest possible.
Relevant Equations:: The solution is k = $\sqrt{2}-1$ and $a_1 = 2 \cdot \sqrt{2}$

I do not have any reasonable attempts at this problem, as I am trying to figure out how one can get the correct answer when we are not given any values. Maybe if some of you sees a mistake here, that implies that the values from the previous example should be used...
A hint to start: How do you express the terms of a geometric series?

And, no, this is not true of all geometric series.

Give it another try and if you are still stuck show us what you've got and we'll give you a hand. We can't really help you until we know where the problem is. (And, of course, as PeroK says, use ## on both sides of the code.)

-Dan

Aristarchus_ and Greg Bernhardt
You should fix the Latex.
It's now fixed.

jim mcnamara and topsquark
A hint to start: How do you express the terms of a geometric series?

And, no, this is not true of all geometric series.

Give it another try and if you are still stuck show us what you've got and we'll give you a hand. We can't really help you until we know where the problem is. (And, of course, as PeroK says, use ## on both sides of the code.)

-Dan
I've updated my progress so far above

##a_1 \cdot (1+k^{2}) = a_1^{2} \cdot k##
But I do not see any connections yet...
that is the equation relatIng ##k## and ##a_1##.

Now you are asked to minimise the sum.

Please post any updates in a new post, not in the original.

topsquark and Aristarchus_
Update:
I basically took the derivative of the sum of the geometric series with respect to k and set it equal to 0. Thus I obtained two possible solutions ##k = 1+- \sqrt{2}##.
Now, from the above relationship between ##a_1 + a_3 = a_1 \cdot a_2## I got the expression ##k = (a_{1} + \sqrt{a_{1}^2 - 4)}/2## and inserting k value in it I got ##a = 2\sqrt{2}##. Is this a valid approach, and are there any corrections?

Mod note: I tried to clean up your LaTeX, but am not sure what I changed it to is what you meant.

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Update:
I basically took the derivative of the sum of the geometric series with respect to k and set it equal to 0. Thus I obtained two possible solutions ##k = 1+- \sqrt{2}##.
Now, from the above relationship between ##a_1 + a_3 = a_1 \cdot a_2## I got the expression ##k = ((a_{1} + \sqrt(a_{1}^(2) - 4))/(2))## and inserting k value in it I got #a = 2\sqrt{2}#. Is this a valid approach, and are there any corrections?
How did you differentiate the sum of the series?

How did you differentiate the sum of the series?
S = a_1 \cdot (k^{n} - 1/ k-1) , this in respect to k and setting n = 1

S = a_1 \cdot (k^{n} - 1/ k-1) , this in respect to k and setting n = 1
Do you mean ##n = \infty##?

What did you do about ##a_1##?

##a_3 = a_1 \cdot k^2## and ##a_2=a_1 \cdot k##
##a_1 + a_3 = a_1 \cdot (1+k^{2})##
##a_1 \cdot a_2 = a_1^{2} \cdot k##
from the problem, it then follows that ##a_1 \cdot (1+k^{2}) = a_1^{2} \cdot k##
But I do not see any connections yet...
Derivative? Just solve for k! It's a quadratic.

-Dan

Do you mean ##n = \infty##?

What did you do about ##a_1##?
##a_1## cancels out when the derivative is set to 0...

Derivative? Just solve for k! It's a quadratic.

-Dan
Which expression and what do you get then?

##a_1## cancels out when the derivative is set to 0...
I don't see how you got ##k = \sqrt 2 - 1## by minimising ##S(k) = \dfrac{a_1}{1-k}##?

Derivative? Just solve for k! It's a quadratic.

-Dan
The problem is to find the minimum value of the sum given the relationship between ##a_1## and ##k##.

topsquark
The problem is to find the minimum value of the sum given the relationship between ##a_1## and ##k##.
I knew that. But for some reason I was thinking once you have k then you look for a minimum. Silly idea, really. Thanks for the catch!

-Dan

I don't see how you got ##k = \sqrt 2 - 1## by minimising ##S(k) = \dfrac{a_1}{1-k}##?
I took the derivative of ##a_1 \cdot \frac{k^{n}-1}{k-1}##. But how do I determine the minimum of ##S(k) = \dfrac{a_1}{1-k}##? What did you have in mind?

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I took the derivative of ##a_1 \cdot \frac{k^{n}-1}{k-1}##. But how do I determine the minimum of ##S(k) = \dfrac{a_1}{1-k}##? What did you have in mind?
That's the formula for a finite series. You have an infinite series, which is simpler.

My point is that ##a_1## depends on ##r##. You need to take that into account.

PS I used ##r## instead of ##k## here. Apologies.

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S = a_1 \cdot (k^{n} - 1/ k-1) , this in respect to k and setting n = 1
Try to be more consistent in your use of LaTeX. The equation above would be ##S = a_1 \frac{k^n - 1}{k - 1}##. Take a look at our tutorial on LaTeX -- there's a link in the lower left corner of the text input area.

Also, your expression (k^n - 1/k - 1) would normally be interpreted as ##k^n - \frac 1 k - 1##, which surely isn't what you meant.

What you need to finish this problem is to use the given information from the problem statement to solve for ##a_1## in terms of k. Once you get the sum of the series entirely as a function of k, it's straightforward to differentiate the sum function and find the value of k that minimizes the sum. There will be two values, but only one of them is applicable, given that this is a convergent geometric series.

BTW, @PeroK is using ##r## for the common ratio, while you are using ##k##.

topsquark and PeroK
But the problem is almost solved already in #1. From the attempt you immediately get ##a_1## as a function of ##k##. Then you can calculate the infinite sum,
$$S(k)=\sum_{n=1}^{\infty} a_n=a_1 \sum_{n=1}^{\infty} k^{n-1},$$
plug in your solution for ##a_1## and minimize the resulting function ##S(k)##.

A maybe better solution...

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