Question

Find the values for x for which the series converges and find the sum of the series (as a function) for those values of x:

Sum (-1)^2 * (2^(n+2))/(3^(2n-1)) * (x+3)^(2n+1)
{n,0, Infinity}

Answer 1

(-1)^2 .... why is that there .... won't that just be 1 ???

Answer 2

can you verify this is correct?

Answer 3

@kreimer20 are you there?

Answer 4

Sorry, I'm trying to work it out atm... But that is how my professor listed it... To me that's just a 1.. but maybe it is wanting you to re-index the problem?

Answer 5

\[\Large \sum_{n=0}^\infty(-1)^2 \cdot \left( 2^{n+2}\over 3^{2n-1} \right)\cdot \left( x+3 \right)^{2n+1}\]

Answer 6

yes
that is the equation

Answer 7

i was wondering if it was supposed to be (-1)^n .... but if it is what it is we can just remove (-1)^2 because it is 1

this looks like a Geometric series ... but you are right we need to re-write the indices .... we can only use n.
... so we can tell what the first and common ratio are.

Answer 8

Correction, n=1 not n=0 for the problem above.

Answer 9

uhh ok ... then we need to make all indices n-1

Answer 10

do reindex first.
Then after that, I assume solve to find what the series converges to. But idk how to fit that into the question the problem is asking

Answer 11

\[\Large a_n= \left( 8\cdot 2^{n-1} \over 3\cdot9^{n-1}\right)\left( x+3 \right)^3\left[ \left( x+3 \right)^2 \right]^{n-1}\]

Answer 12

why are there now two (x+3)'s?

Answer 13

because we can only use n-1 ...(or a number) ... as power. you should be able to that this is equivalent to the original

Answer 14

\[\large \implies a_n=\left( 8\left( x+3 \right)^3\over3 \right)\times \left[ 2\left( x+3 \right)^2\over9 \right]^{n-1}\]

^^^ this now looks like the n-th term of a Geometric Series\[\large a_n=a \cdot r^{n-1}\]

Answer 15

did you follow?

Answer 16

fuuuu**ck... It was (-1)^n... I was rewriting it into my notebook when i realized I had read it wrong... twice...

Answer 17

I understand what you have done so far tho

Answer 18

that's easy enough to include... just want to also make sure the sum is from 1 to infty.

....(and not 0 to infty like you originally said)

Answer 19

\[\large \implies a_n=\left(- 8\left( x+3 \right)^3\over3 \right)\times \left[- 2\left( x+3 \right)^2\over9 \right]^{n-1}\]

Answer 20

yes, 1 to infinity

Answer 21

A geometric series converges only when \[-1<r<1\]

Answer 22

so it doesn't converge?

Answer 23

so the nth term you get, is that filling in 0 for n and rewritting the equation to reflect that?

Answer 24

so, \[\large -1<-{2\left( x+3 \right)^2\over9 }<1\]

Answer 25

we have to find the value of x for which it will converge

Answer 26

Part 1: \[\large -1<-{2\left( x+3 \right)^2\over9 }\\ \implies 9> 2\left( x+3 \right)^2\\ \implies 9> 2x^2+12x+18 \\ \implies 2x^2+12x+9<0\\ \implies \color{blue}{{-6-3\sqrt2\over2}<x<{-6+3\sqrt2\over2}}\]

Answer 27

Part 2:\[-{2\left( x+3 \right)^2\over9 }<1\\ \implies \large\left( x+3 \right)^2>-\frac92\]... this is true for all real numbers ...(squares are always >=0)




So the series converges when:\[\large \color{blue}{{-6-3\sqrt2\over2}<x<{-6+3\sqrt2\over2}}\]

Answer 28

when a Geometric Series converges the sum is giver by the formula: \[\Large S={a \over1-r}\]

\[\Large{=\frac{- 8\left( x+3 \right)^3\over3 }{ 1-{- 2\left( x+3 \right)^2\over9}}\\=\color{green}{-24\left( x+3 \right)^3 \over 2x^2+12x+27}}\]

Answer 29

I'm not really understanding how you got from the reindex to the a-sub-n term with the (-1)^n-1

Answer 30

\[\Large a_n=(-1)^n \cdot \left( 2^{n+2}\over 3^{2n-1} \right)\cdot \left( x+3 \right)^{2n+1}\]
now, \[(-1)^n=(-1)\times(-1)^{n-1}\]


since we did the first reindex without (-1)^n term present ... i just multiplied -1 to a and to r in the first calculation to compensate

Answer 31

okay you might have to walk me through that whole thing... because I saw how the 2 (x+3)'s were brought down into one (x+3); I was able to do it. I'm having a really tough time getting the (-1)^n and the 2^(n+1)/3^(2n-3) to come down into the condensed form you have written out.

Answer 32

yeah let me just re-write all the steps: there is a little bit of confusion because of the initial error:

we start with:\[\Large a_n=(-1)^n \cdot \left( 2^{n+2}\over 3^{2n-1} \right)\cdot \left( x+3 \right)^{2n+1}\]


and we want to rewrite using only n-1 power

Answer 33

n = (n-1)+1
n+2 = (n-1)+3
2n-1 = 2(n-1)+1
2n+1 = 2(n-1)+3

we get:\[\large a_n= (-1)\cdot(-1)^{n-1}\times \left( 8\cdot 2^{n-1} \over 3\cdot9^{n-1}\right)\times\left( x+3 \right)^3\left[ \left( x+3 \right)^2 \right]^{n-1}\]

Answer 34

next we separate everything that is not raised to the (n-1) power and everything that is.

Answer 35

\[\large \implies a_n=\left((-1) 8\left( x+3 \right)^3\over3 \right)\times \left[(-1) 2\left( x+3 \right)^2\over9 \right]^{n-1}\]

Answer 36

and that's it !

Answer 37

what happened to the (-1)^(n-1)?

Answer 38

every thing that was raised to the (n-1) power was moved inside the brackets [ ] in the right.

Answer 39

that's why there is a (-1) inside the [ ]

Answer 40

Thank you so very much. I finally understood it. Good teacher and thank you for your patience!