# Geometric Series of nr^n

## 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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etotheipi
Gold Member
2019 Award
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.

• PeroK
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)

etotheipi
Gold Member
2019 Award
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.

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

etotheipi
Gold Member
2019 Award
It's certainly not the same question, because you have ignored all of the terms except the final line...  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) ?

etotheipi
Gold Member
2019 Award
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.

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

PeroK
Homework Helper
Gold Member
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##?

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)

etotheipi
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2019 Award
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!

Last edited:
PeroK
Homework Helper
Gold Member
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##.

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) ?

etotheipi
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2019 Award
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.

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)

etotheipi
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2019 Award
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!

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.

etotheipi
Gold Member
2019 Award
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?

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)

• etotheipi
etotheipi
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2019 Award
So do you see how the proof concludes?

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?

etotheipi
Gold Member
2019 Award
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.

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 etotheipi
Gold Member
2019 Award
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}##

• jisbon