Proving the Geometric Series with Variable Coefficients: A Scientific Approach

In summary: No. Sometimes it can be helpful to try different numbers, but it's not necessary.No. Sometimes it can be helpful to try different numbers, but it's not necessary.
  • #1
jisbon
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Homework Statement
Show that summation of r^n+2r^n+3r^n+...nr^n is less than r/(1-r^3) for n more than one.
Relevant Equations
-
So this seems to be a geometric Series, but with the coefficients in front, how do I exactly go about proving this?
Thanks
 
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  • #2
jisbon said:
Show that summation of r^n+2r^n+3r^n+...nr^n is less than r/(1-r^3) for n more than one.

Are you sure you did write the question down correctly? As it stands,$$r^n + 2r^n + 3r^n + \dots + nr^n = r^n (1+2+3+ \dots + n) = \frac{n(n+1)r^n}{2}$$which doesn't appear to be what is intended.
 
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  • #3
etotheipi said:
Are you sure you did write the question down correctly? As it stands,$$r^n + 2r^n + 3r^n + \dots + nr^n = r^n (1+2+3+ \dots + n) = \frac{n(n+1)r^n}{2}$$which doesn't appear to be what is intended.
It is correct, but how does one prove that it is lesser than r/(1-r^3)
 
  • #4
jisbon said:
It is correct, but how does one prove that it is lesser than r/(1-r^3)

It's not; that's why I presumed the question was written down incorrectly, or otherwise that the question itself is wrong.
 
  • #5
etotheipi said:
It's not; that's why I presumed the question was written down incorrectly, or otherwise that the question itself is wrong.
I've DMed you the question, and it seems to be exactly what I written down here
 
  • #6
It's certainly not the same question, because you have ignored all of the terms except the final line... :wink:

1598951041851.png
 
  • #7
etotheipi said:
It's certainly not the same question:

View attachment 268701
Isn't this proving the summation of r^n + 2r^n + ... is less than r/(1-r^3) ?
 
  • #8
jisbon said:
Isn't this proving the summation of r^n + 2r^n + ... is less than r/(1-r^3) ?

No, because that is a necessary condition, but not a sufficient condition!

Also note that we now know the key bit of information, that ##0 < r < 1##! Without this, the question would not make any sense.
 
  • #9
etotheipi said:
No, because that is a necessary condition, but not a sufficient condition!

Also note that we now know the key bit of information, that ##0 < r < 1##! Without this, the question would not make any sense.
Hmm okay, so how do I start about proving this? I tried dividing the hint given by (1-r) again to make both similar, but I can't seem to get the gist of it
 
  • #10
jisbon said:
Hmm okay, so how do I start about proving this? I tried dividing the hint given by (1-r) again to make both similar, but I can't seem to get the gist of it
What about ##r = 0.5##, ##n = 2## or ##n = 3## or ##n = 4##?
 
  • #11
PeroK said:
What about ##r = 0.5##, ##n = 2## or ##n = 3## or ##n = 4##?
As in computing them?
With r=0.5, it would seem impossible to compute out the equations, but it does converge eventually. Could you kindly advice me how to compute? (I was thinking r>1 all along)
 
  • #12
I would start with noticing that $$\begin{align*}

S = &+ 1(r + r^2 + r^3 + \dots + r^n) \\

&+ 2(r^2 + r^3 + r^4 + \dots + r^n) \\

&+ 3(r^3 + r^4 + r^5 + \dots + r^n) \\

&+ \dots \\

&+ nr^n
\end{align*}$$Notice that this can be simplified using the formulae for geometric progressions to$$S = \frac{r(r^n -1)}{r-1} + 2\frac{r^2(r^{n-1} -1)}{r-1} + \dots + n\frac{r^n(r^1 - 1)}{r-1}$$ $$S = \frac{1}{r-1} \left[ (r^{n+1} - r) + (2r^{n+1} - 2r^2) + (3r^{n+1} - 3r^3) + \dots + (nr^{n+1} - nr^n) \right]$$ $$S = \frac{1}{r-1} \left[ r^{n+1}(1+2+3+ \dots + n) - (r + 2r^2 + 3r^3 + \dots + nr^n) \right]$$ $$S = \frac{n(n+1)}{2} \frac{r^{n+1}}{r-1} - \frac{(r + 2r^2 + 3r^3 + \dots + nr^n)}{r-1}$$Note that for ##0 < r < 1##, the first term is less than zero... play around with this, and see if you can get it in the right form!
 
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  • #13
jisbon said:
As in computing them?
With r=0.5, it would seem impossible to compute out the equations, but it does converge eventually. Could you kindly advice me how to compute? (I was thinking r>1 all along)
If ##r > 1##, then the right hand side is negative: ##1- r^3## is negative.

There's no convergence, as it's a fixed ##n##.
 
  • #14
etotheipi said:
I would start with noticing that $$\begin{align*}

S = &+ 1(r + r^2 + r^3 + \dots + r^n) \\

&+ 2(r^2 + r^3 + r^4 + \dots + r^n) \\

&+ 3(r^3 + r^4 + r^5 + \dots + r^n) \\

&+ \dots \\

&+ n(r^n)
\end{align*}$$Notice that this can be simplified using the formulae for geometric progressions to$$S = \frac{r(r^n -1)}{r-1} + 2\frac{r^2(r^{n-1} -1)}{r-1} + \dots + n\frac{r^n(r^1 - 1)}{r-1}$$ $$S = \frac{1}{r-1} \left[ (r^{n+1} - r) + (2r^{n+1} - 2r^2) + (3r^{n+1} - 3r^3) + \dots + (nr^{n+1} - nr^n) \right]$$ $$S = \frac{1}{r-1} \left[ r^{n+1}(1+2+3+ \dots + n) - (r + 2r^2 + 3r^3 + \dots + nr^n) \right]$$ $$S = \frac{n(n+1)}{2} \frac{r^{n+1}}{r-1} - \frac{(r + 2r^2 + 3r^3 + \dots + nr^n)}{r-1}$$Note that for ##0 < r < 1##, the first term is less than zero... play around with this, and see if you can get it in the right form!
Hmm, I don't seem to see the relation at the start. How does one get +1(r+r^2+r^3+...r^n) ?
 
  • #15
jisbon said:
Hmm, I don't seem to see the relation at the start. How does one get +1(r+r^2+r^3+...r^n) ?

I just listed out all of the terms. ##S## is equal to the whole sum, not just that line.
 
  • #16
etotheipi said:
I just listed out all of the terms. ##S## is equal to the whole sum, not just that line.
Yes I understand you have listed all the terms out with also +2(...) and +3(...), but I can't seem to see why it is listed as so though. (Can't seem to get the pattern)
 
  • #17
jisbon said:
Yes I understand you have listed all the terms out with also +2(...) and +3(...), but I can't seem to see why it is listed as so though. (Can't seem to get the pattern)

Each line in that sum is a column in the original problem statement!
 
  • #18
etotheipi said:
Each line in that sum is a column in the original problem statement!
Oh no, I think I know why. I might have misread the question (thinking that they want the sum of 1 line instead of all the lines :/ Will reread the question now, will reply back if I have any additional queries. Thanks for the correction.
 
  • #19
Rules Violation: Member warned about providing too much help
To give a further hint, we deduced that$$S = \frac{n(n+1)}{2} \frac{r^{n+1}}{r-1} - \frac{(r + 2r^2 + 3r^3 + \dots + nr^n)}{r-1}$$Add ##\frac{r}{(r-1)^3}## to both sides,$$\begin{align*}S + \frac{r}{(r-1)^3} &= \frac{n(n+1)}{2} \frac{r^{n+1}}{r-1} + \frac{1}{(r-1)} \left[ \frac{r}{(r-1)^2} - (r + 2r^2 + 3r^3 + \dots + nr^n) \right] \\ \\

&=\frac{n(n+1)}{2} \frac{r^{n+1}}{r-1} + \frac{1}{(r-1)} \left[ \frac{r}{(r-1)^2} - \sum_{i=1}^n ir^i \right]

\end{align*}$$What can you say about the RHS?
 
  • #20
etotheipi said:
To give a further hint, we deduced that$$S = \frac{n(n+1)}{2} \frac{r^{n+1}}{r-1} - \frac{(r + 2r^2 + 3r^3 + \dots + nr^n)}{r-1}$$Add ##\frac{r}{(r-1)^3}## to both sides,$$\begin{align*}S + \frac{r}{(r-1)^3} &= \frac{n(n+1)}{2} \frac{r^{n+1}}{r-1} + \frac{1}{(r-1)} \left[ \frac{r}{(r-1)^2} - (r + 2r^2 + 3r^3 + \dots + nr^n) \right] \\ \\

&=\frac{n(n+1)}{2} \frac{r^{n+1}}{r-1} + \frac{1}{(r-1)} \left[ \frac{r}{(r-1)^2} - \sum_{i=1}^n ir^i \right]

\end{align*}$$What can you say about the RHS?
RHS:
(less than 0) + (negative number)(positive-negative) =
(less than 0) + (negative number)(positive) =
(less than 0) + (negative number) =
(negative number)
 
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  • #21
So do you see how the proof concludes?
 
  • #22
etotheipi said:
So do you see how the proof concludes?
So if S + r/(r-1)^3 is negative, minus r/(r-1)^3 on both sides, I get S = some negative number - r/(r-1)^3...
How is S less than - r/(r-1)^3 ?
Also, I was taught that the formula for GP is (1-r^(n+1)) / 1-r, can this still apply?
 
  • #23
jisbon said:
So if S + r/(r-1)^3 is negative, minus r/(r-1)^3 on both sides, I get S = some negative number - r/(r-1)^3...
How is S less than - r/(r-1)^3 ?

Essentially yes; if the RHS is negative, then so is the LHS (they're equal!). We can write$$S + \frac{r}{(r-1)^3} < 0$$ $$S < \frac{-r}{(r-1)^3} = \frac{r}{(1-r)^3}$$which is what we set out to prove.

jisbon said:
Also, I was taught that the formula for GP is (1-r^(n+1)) / 1-r, can this still apply?

I don't think that's right. What do you mean by 'formula for GP'? The sum of the first ##n## terms of a GP with first term ##a## and common ratio ##r## is$$S = \frac{a(1-r^n)}{1-r} = \frac{a(r^n - 1)}{r-1}$$but that's not what you wrote.
 
  • #24
etotheipi said:
Essentially yes; if the RHS is negative, then so is the LHS (they're equal!). We can write$$S + \frac{r}{(r-1)^3} < 0$$ $$S < \frac{-r}{(r-1)^3} = \frac{r}{(1-r)^3}$$which is what we set out to prove.
I don't think that's right. What do you mean by 'formula for GP'? The sum of the first ##n## terms of a GP with first term ##a## and common ratio ##r## is$$S = \frac{a(1-r^n)}{1-r} = \frac{a(r^n - 1)}{r-1}$$but that's not what you wrote.
Understood your first portion, thanks a lot :)
Regarding GP, what I meant was the formula for geometric series sorry. I was taught that
1598958821468.png
 
  • #25
jisbon said:
Understood your first portion, thanks a lot :)
Regarding GP, what I meant was the formula for geometric series sorry. I was taught thatView attachment 268703

That's also fine. In my experience it's more common to start counting from ##u_1##, i.e. to consider ##S = \sum_{j = 1}^n ar^{j-1} = \frac{a(1-r^n)}{1-r}##, but there's nothing wrong at all with shifting ##k = j-1## to find ##S = \sum_{k = 0}^{n-1} ar^k = \frac{a(1-r^n)}{1-r}##
 
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  • #26
I did it quite similarly to @etotheipi. My starting point was to make the sum vertically (your picture).
$$S=\sum_{j=1}^nj\left(\sum_{i=j}^nr^i\right)$$
I know how to compute a geometric sum from ##i=0## up to some number, but here I want to start from some arbitrary index, so my plan is to make two sums starting from ##i=0##, but with one ending at ##i=n##, and the other at ##i=j-1##, then subtract both
$$\begin{align*}
S&=\sum_{j=1}^nj\left(\sum_{i=0}^{n}r^i-\sum_{i=0}^{j-1}r^i\right)\\
&=\sum_{j=1}^nj\left(\frac{1-r^{n+1}}{1-r}-\frac{1-r^j}{1-r}\right)\\
&=\sum_{j=1}^nj\left(\frac{r^j-r^{n+1}}{1-r}\right)\\
&=\left(\sum_{j=1}^n\frac{jr^j}{1-r}\right)-\left(\frac{r^{n+1}}{1-r}\sum_{j=1}^nj\right)
\end{align*}$$
Since ##0<r<1##, I know that the second parenthesis is positive because ##1-r>0##, and so
$$S=\left(\sum_{j=1}^n\frac{jr^j}{1-r}\right)-\left(\frac{r^{n+1}}{1-r}\sum_{j=1}^nj\right)<\sum_{j=1}^n\frac{jr^j}{1-r}$$
If you remove something from a positive number, then, of course, that positive number is bigger than the result of that operation.
Using the hint, we get
$$\sum_{j=1}^njr^j<\frac{r}{(1-r)^2}\Leftrightarrow S<\frac{1}{1-r}\sum_{j=1}^njr^j=\sum_{j=1}^n\frac{jr^j}{1-r}<\frac{r}{(1-r)^3}$$
 
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  • #27
Can someone tell me why I rewrote what @etotheipi posted? :oldconfused:
I'm amazed by myself. o_O
 
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  • #28
archaic said:
Can someone tell me why I rewrote what @etotheipi posted? :oldconfused:

I think yours is an improvement; the sigma notation makes everything much more compact.
 
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  • #29
etotheipi said:
Nice! There is a very tiny typo, that the first formula should I think be$$S=\sum_{j=1}^nj\left(\sum_{i\geq j}^nr^i\right)$$

I think yours is an improvement; the sigma notation makes everything much more compact.
I think that we can write ##i=j##. I did a quick google search and have found someone else using it like that.
Not very convincing, but yeah. :blushing:
 
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  • #30
archaic said:
I think that we can write ##i=j##. I did a quick google search and have found someone else using it like that.
Not very convincing, but yeah. :blushing:

Sorry, no you're right. I missed the upper index! :doh:
 
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1. What is a geometric series with variable coefficients?

A geometric series with variable coefficients is a mathematical sequence in which each term is multiplied by a different number, known as the coefficient. The series follows a specific pattern in which each term is a multiple of the previous term, creating a geometric progression.

2. How is the geometric series with variable coefficients proven?

The geometric series with variable coefficients can be proven using a scientific approach, which involves analyzing the series using mathematical principles and equations. This includes identifying the common ratio between terms, using the sum of a geometric series formula, and using mathematical induction to verify the proof.

3. What are the applications of proving the geometric series with variable coefficients?

The proof of the geometric series with variable coefficients has various applications in mathematics, physics, and engineering. It is used to solve problems involving compound interest, population growth, and radioactive decay. It also helps in understanding the concept of limits and convergence in calculus.

4. What are the challenges in proving the geometric series with variable coefficients?

One of the main challenges in proving the geometric series with variable coefficients is finding the common ratio between terms, as it may not always be obvious. Additionally, the proof may involve complex mathematical equations and requires a thorough understanding of geometric series and mathematical induction.

5. How does the proof of the geometric series with variable coefficients contribute to scientific knowledge?

The proof of the geometric series with variable coefficients contributes to scientific knowledge by providing a deeper understanding of the properties and behavior of geometric series. It also serves as a basis for solving more complex problems in mathematics and other scientific fields, further advancing our understanding of the world around us.

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