Geometric Series of nr^n

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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
etotheipi
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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
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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
etotheipi
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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
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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
etotheipi
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It's certainly not the same question, because you have ignored all of the terms except the final line... :wink:

1598951041851.png
 
  • #8
etotheipi
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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
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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
PeroK
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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
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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
etotheipi
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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
PeroK
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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
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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
etotheipi
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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
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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
etotheipi
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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
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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
etotheipi
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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
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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
etotheipi
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So do you see how the proof concludes?
 
  • #22
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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
etotheipi
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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.

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
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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
etotheipi
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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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