Geometric Series: x_n = 1 + 2/9 + ... + (2/9)^n; find limit as n -> infinity

[Kemikeren]

I have the following problem:

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Consider the sequence, \(\displaystyle x_1,\, x_2,\, ...,\, x_n,\, ...,\) defined by:

. . . . .\(\displaystyle x_1\, =\, 1\, +\, \dfrac{2}{9}\)

and

. . . . .\(\displaystyle x_n\, =\, 1\, +\, \dfrac{2}{9}\, +\, ...\, +\, \left(\dfrac{2}{9}\right)^n\)

for \(\displaystyle n\, =\, 2,\, 3\, ...\)

Determine the limit:

. . . . .\(\displaystyle \displaystyle x\, =\, \lim_{n \rightarrow \infty}\, x_n\)

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I have tried to solve this problem but without luck. How do I proceed?

 

[Harry_the_cat]

I have the following problem:

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Consider the sequence, \(\displaystyle x_1,\, x_2,\, ...,\, x_n,\, ...,\) defined by:

. . . . .\(\displaystyle x_1\, =\, 1\, +\, \dfrac{2}{9}\)

and

. . . . .\(\displaystyle x_n\, =\, 1\, +\, \dfrac{2}{9}\, +\, ...\, +\, \left(\dfrac{2}{9}\right)^n\)

for \(\displaystyle n\, =\, 2,\, 3\, ...\)

Determine the limit:

. . . . .\(\displaystyle \displaystyle x\, =\, \lim_{n \rightarrow \infty}\, x_n\)

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I have tried to solve this problem but without luck. How do I proceed?

This is the same as finding the sum of an infinite series. a=1, r=2/9 (which is < 1). Do you know how to do that?

 

[Kemikeren]

This is the same as finding the sum of an infinite series. a=1, r=2/9 (which is < 1). Do you know how to do that?

Well I know that S = a/1-r for |r|<1 - But when i try to calculate this, I do not seem to get the right result. So it seems like I am inserting the wrong values.

 

[HallsofIvy]

Well I know that S = a/1-r for |r|<1 - But when i try to calculate this, I do not seem to get the right result. So it seems like I am inserting the wrong values.

Well, in the problem here a= 1 and r= 2/9. What do you get when you put those in?


(And please don't tell me you are using a calculator to do this easy computation and entering (a/1)- r rather than a/(1- r)!)

 

[Steven G]

Well I know that S = a/1-r for |r|<1 - But when i try to calculate this, I do not seem to get the right result. So it seems like I am inserting the wrong values.

You are not getting the correct result because S\(\displaystyle \neq\) a-1/r, rather S = a/(r-1). Please try again and do not use a calculator!

 

[Kemikeren]

You are not getting the correct result because S\(\displaystyle \neq\) a-1/r, rather S = a/(r-1). Please try again and do not use a calculator!

I did not use a calculator (of course), but since n = 2,3... I thought that I had to say (2/9)^2 as "r".

Thank you for your help, I suppose the result is 9/7

 

[Harry_the_cat]

I did not use a calculator (of course), but since n = 2,3... I thought that I had to say (2/9)^2 as "r".

Thank you for your help, I suppose the result is 9/7

Yes 9/7 is correct.
The "n = 2, 3, …" refers to the definition of the function. The first term (ie when n=1) is defined separately, as is usually the case for iterative functions.

 

[Kemikeren]

Yes 9/7 is correct.
The "n = 2, 3, …" refers to the definition of the function. The first term (ie when n=1) is defined separately, as is usually the case for iterative functions.

Thank you so much for your help !

 

[Steven G]

I did not use a calculator (of course), but since n = 2,3... I thought that I had to say (2/9)^2 as "r".

Thank you for your help, I suppose the result is 9/7

If r = (2/9)^2, then the series would have been (2/9)^2 + (2/9)^4 + (2/9)^6 + (2/9)^8 +....

Now if you felt that the series was (2/9)^2 + (2/9)^3 + (2/9)^4 + (2/9)^5 +...., then you should have computed (2/9)^0 + (2/9)^1 + (2/9)^2 + (2/9)^3 + (2/9)^4 + (2/9)^5+... -((2/9)^0 + (2/9)^1) = (2/9)^0 + (2/9)^1 + (2/9)^2 + (2/9)^3 + (2/9)^4 + (2/9)^5+... -1 2/9.