Question

Evaluate \(\sum_{n=0}^\infty x^{n} (n^{2} + n)\) assuming the series converges

Answer 1

x has to be between 0 and 1 then ..... right?

Answer 2

yes.. |x| < 1 for convergence

Answer 3

and a power function (not technical term for it) grows faster than a poly right? so it would at least converge

Answer 4

0r^0 + 0r^0
1r^1 + 1r^1
4r^2 + 2r^2
9r^2 + 3r^3

the right side we did earlier converges to 1/(1-r)^2

Answer 5

Ahh we can use that i guess
earlier we did \(\sum\limits_{n=0}^{\infty} nx^n = \dfrac{x}{(1-x)^2}\)

Answer 6

*x in numerator

Answer 7

. = 1 + r + 4r^2 +9r^3 +16r^4 + ... + (n-1)^2 r^(n-1)
-r. = - r - r^2 - 4r^3 - 9r^4 - ... - (n-2)^2 r^(n-1) -(n-1)^2r^n
-----------------------------------------------------------
(1-r).= 1 +3r^2 +5r^3 +7r^4 + .... + (2n-3) r^(n-1) - (n-1)^2 r^n

and the next one of course reduces to constants

Answer 8

(1-r).= 1 +3r^2 +5r^3 +7r^4 + .... + (2n-3) r^(n-1) - (n-1)^2 r^n
-r(1-r).= -r -3r^3 -5r^4 - .... - (2n-3) r^n
+ (n-1)^2 r^(n+1)
---------------------------------------------------------------
(1-r)^2, = 1-r+3r^2 +2r^3 +2r^4 + ... + 2r^(n-1) -[(n^2-2] r^n
+ (n-1)^2 r^(n+1)

anohter -r seems to be able to get us there. with just a few term left over

Answer 9

that works nicely, the sum n^2r^n doesn't start with 1
```
. = r + 4r^2 +9r^3 +16r^4 + ... + (n-1)^2 r^(n-1)
-r. = - r - r^2 - 4r^3 - 9r^4 - ... - (n-2)^2 r^(n-1) -(n-1)^2r^n
-----------------------------------------------------------
(1-r).= 3r^2 +5r^3 +7r^4 + .... + (2n-3) r^(n-1) - (n-1)^2 r^n
```
and the next one of course reduces to constants

Answer 10

n^2 r^n by the wolf gets us reduced to r^2+r/(1-r)^3

but id have to work it on paper to keep it organized better :)

Answer 11

nooo, what we had will do haha we're mostly there :)

Answer 12

0r^0=0 , yeah :)

Answer 13

x = r in this thread

Answer 14

*then SUM x^n (n^2+n) evaluates to

(r^2+r)/(1-r)^3 + r/(1-r)^2 = 2r/(1-r)^3

Answer 15

thought right term was 1/(1-r)^2 or am i picking a typo?

Answer 16

\[\dfrac{1}{1-r} = \sum\limits_{n=0}^{\infty} r^n \]
differentiating both sides gives
\[\dfrac{1}{(1-r)^2} = \sum\limits_{n=0}^{\infty} nr^{\color{red}{n-1}} \]


we need to multiply \(r\) both sides to fix that exponent on right hand side right

Answer 17

right, got it :)

Answer 18

ive always wondered if it was just convention, but derivative of a summation index adds 1

Answer 19

\[y=\sum_0a_nx^n\]
\[y'=\sum_1a_nnx^{n-1}\]

Answer 20

Ahh right but there is no harm in starting index at 0 right, we will just get the first term 0
or will it shift the \(a_i\) 's

Answer 21

\[\sum\limits_{n=0}^{\infty} nr^{n} = \sum\limits_{n=1}^{\infty} nr^{n} \]

Answer 22

that seems fair for the first derivative :)

Answer 23

we get more leading 0's as we differentiate more times
other than that.. hmm is there any harm

Answer 24

kx^0 if we just powe rule it is: 0k/x = 0

Answer 25

i see.. that 0 coefficient still makes it 0

Answer 26

but using power rule on x^0 doesn't look correct in series yeah

Answer 27

that might be the reason we explicitly change the start of index when differentiating hmm

Answer 28

something like that yes. helps avoid some technical issues.

Answer 29

\[\frac{a_0}{1-r}=\sum_0^{inf}a_n r^n\]
\[\frac{a_0}{1-r}=a_0+\sum_1^{inf}a_n r^n\]
\[-\frac{a_0}{(1-r)^2}=0+\sum_1^{inf}a_nn r^{n-1}\]

Answer 30

\[-\frac{a_0}{(1-r)^2}=\sum_0^{inf}a_{n+1}(n+1) r^{n}\]

Answer 31

do i have an error?

Answer 32

the negative is an error ... that gets me everytime ..

Answer 33

i think \(a_n \) needs to be a constant for using that infinite geometric series formula a1/(1-r)

Answer 34

\[\frac{a_0}{1-r}=\sum_0^{inf}a_n r^n\]

this works oly if \(a_n = a_0\) for all \(n\)

Answer 35

true true, but at 315am it all looks the same :)

Answer 36

lmao
i have another way to work this problem but may be il post it later as i don't want to keep u from sleeping lol

Answer 37

very well you have a good night/morning wherever you are ;)

Answer 38

gnite!