Proving the Geometric Series with Variable Coefficients: A Scientific Approach

[jisbon]

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

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So this seems to be a geometric Series, but with the coefficients in front, how do I exactly go about proving this?
Thanks

 

[etotheipi]

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.

 

[jisbon]

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]

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.

 

[jisbon]

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]

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

 

[jisbon]

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]

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.

 

[jisbon]

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]

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 $n = 4$?

 

[jisbon]

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]

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!

 

[PeroK]

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 $n$.

 

[jisbon]

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]

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.

 

[jisbon]

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]

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!

 

[jisbon]

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]

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?

 

[jisbon]

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]

So do you see how the proof concludes?

 

[jisbon]

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]

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.

 

[jisbon]

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]

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 $S = \sum_{k = 0}^{n-1} ar^k = \frac{a(1-r^n)}{1-r}$

 

[archaic]

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
$$\sum_{j=1}^njr^j<\frac{r}{(1-r)^2}\Leftrightarrow S<\frac{1}{1-r}\sum_{j=1}^njr^j=\sum_{j=1}^n\frac{jr^j}{1-r}<\frac{r}{(1-r)^3}$$

 

[archaic]

Can someone tell me why I rewrote what @etotheipi posted?
I'm amazed by myself.

 

[etotheipi]

Can someone tell me why I rewrote what @etotheipi posted?

I think yours is an improvement; the sigma notation makes everything much more compact.

 

[archaic]

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.

 

[etotheipi]

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.

Sorry, no you're right. I missed the upper index!