Example: Sum x=1 to oo (5x^3+12x^2-10x+123)/(13^x).
I ask because it’s straightforward but I have found only limited special cases.

https://www.idealminischool.ca/idealpedia/index.php/Infinite_Geometric_Series#:~:text=The%20general%20formula%20for%20finding,r%20is%20the%20common%20ratio.

C-B

On Monday, October 17, 2022 at 6:49:39 AM UTC-7, Charlie-Boo wrote:
> Example: Sum x=1 to oo (5x^3+12x^2-10x+123)/(13^x).
> I ask because it’s straightforward but I have found only limited special cases.
>
> https://www.idealminischool.ca/idealpedia/index.php/Infinite_Geometric_Series#:~:text=The%20general%20formula%20for%20finding,r%20is%20the%20common%20ratio.
>
> C-B

Sizes of infinity of the continuum hypothesis are infinities getting contained.

On Monday, October 17, 2022 at 9:49:39 AM UTC-4, Charlie-Boo wrote:
> Example: Sum x=1 to oo (5x^3+12x^2-10x+123)/(13^x).
> I ask because it’s straightforward but I have found only limited special cases.
>
> https://www.idealminischool.ca/idealpedia/index.php/Infinite_Geometric_Series#:~:text=The%20general%20formula%20for%20finding,r%20is%20the%20common%20ratio.
>
> C-B

It is usual to use 'n' as an index of summation, rather than 'x'. Assuming that my algebra is correct, use r = 1/13 in your example

sum (n=0 to infinity) r^n = 1/(1-r) for |r| < 1

sum (n=0 to infinity) n r^n = r/(1-r)^2 = 1/(1-r)^2-1/(1-r) for |r| < 1, via differentiation and multiplying by r

sum (n=0 to infinity) n^2 r^n = 2r/(1-r)^3-r/(1-r)^2 = 2/(1-r)^3-3/(1-r)^2+1/(1-r) for |r| < 1, via differentiation and multiplying by r, etc.

On Monday, October 17, 2022 at 8:27:13 PM UTC-4, Tim Norfolk wrote:
> On Monday, October 17, 2022 at 9:49:39 AM UTC-4, Charlie-Boo wrote:
> > Example: Sum x=1 to oo (5x^3+12x^2-10x+123)/(13^x).
> > I ask because it’s straightforward but I have found only limited special cases.
> >
> > https://www.idealminischool.ca/idealpedia/index.php/Infinite_Geometric_Series#:~:text=The%20general%20formula%20for%20finding,r%20is%20the%20common%20ratio.
> >
> > C-B
> It is usual to use 'n' as an index of summation, rather than 'x'. Assuming that my algebra is correct, use r = 1/13 in your example
>
> sum (n=0 to infinity) r^n = 1/(1-r) for |r| < 1
>
> sum (n=0 to infinity) n r^n = r/(1-r)^2 = 1/(1-r)^2-1/(1-r) for |r| < 1, via differentiation and multiplying by r
>
> sum (n=0 to infinity) n^2 r^n = 2r/(1-r)^3-r/(1-r)^2 = 2/(1-r)^3-3/(1-r)^2+1/(1-r) for |r| < 1, via differentiation and multiplying by r, etc.

Thanks. I was actually looking for an online reference to shorten my reply to a tweet.
https://twitter.com/OnuncuSoru/status/1581650366906540032?s=20&t=GjzGq_ss-U6yd33qCJwFgQ

p = phi

1 + 3/p2 + 5/p4 + 7/p6 + . . . = A L1

p2 + 3 + 5/p2 + 7/p4 + . . . = p2A L2 p2 * L1

p2 + 2 + 2/p2 + 2/p4 + 2/p6 . . . = p2A - A L3 L2 - L1

p4 + 2p2 + 2 + 2/p2 + 2/p4+ 2/p6 + . . . = p4A - p2A L4 p2 * L3

p4 + 2p2 = p4A - 2p2A + A L5 L4 - L3

A*(p4-2p2 +1) = p4 + 2p2

A = (p4 + 2p2) / (p4-2p2 +1)

then solve for a, b, c.