Convergent Series

tingyuau · **Joined:** Wed, 5 Oct 2011 21:02:24 UTC **Posts:** 25

1. The first term in a geometric series is x and the common ratio is x/(1+x). Find the set of values of x for which the series is convergent.

2.The first term in a geometric series is (x-9)/(x+5) and the common ratio is (x+5)/(x-9), where x =/= -5, x=/= 9. For what set of values of x are all the terms of the series positive? Find the set of values of x for which the series is convergent.

I really need some help in doing these, thanks very much!

Shadow · **Posted:** Wed, 5 Oct 2011 21:18:05 UTC

tingyuau wrote:

1. The first term in a geometric series is x and the common ratio is x/(1+x). Find the set of values of x for which the series is convergent.

2.The first term in a geometric series is (x-9)/(x+5) and the common ratio is (x+5)/(x-9), where x =/= -5, x=/= 9. For what set of values of x are all the terms of the series positive? Find the set of values of x for which the series is convergent.

I really need some help in doing these, thanks very much!

You know that an (infinite) geometric series converges exactly when the common ratio has absolute value less than 1. So $\left|{x\over 1+x}\right|<1$ i.e. $|x|<|1+x|\iff x^2<(1+x)^2\iff 0<1+2x\iff -{1\over 2}<x$

You know that the common ratio must be positive for the whole thing to be positive, so ${x+5\over x-9}>0$ , so x>9

or

is necessary. Also, the first term must be positive, but it is just the reciprocal of the common ratio, so it is positive when the common ratio is, so you get: $x\in (-\infty, -5)\cup (9,\infty)$

For the second part, again you need the common ratio to be less than 1, so do the same procedure as I did for the first one, clearing the denominator, squaring, and solving.

tingyuau · **Joined:** Wed, 5 Oct 2011 21:02:24 UTC **Posts:** 25

Shadow wrote:

tingyuau wrote:

1. The first term in a geometric series is x and the common ratio is x/(1+x). Find the set of values of x for which the series is convergent.

2.The first term in a geometric series is (x-9)/(x+5) and the common ratio is (x+5)/(x-9), where x =/= -5, x=/= 9. For what set of values of x are all the terms of the series positive? Find the set of values of x for which the series is convergent.

I really need some help in doing these, thanks very much!

You know that an (infinite) geometric series converges exactly when the common ratio has absolute value less than 1. So $\left|{x\over 1+x}\right|<1$ i.e. $|x|<|1+x|\iff x^2<(1+x)^2\iff 0<1+2x\iff -{1\over 2}<x$

You know that the common ratio must be positive for the whole thing to be positive, so ${x+5\over x-9}>0$ , so x>9

or

is necessary. Also, the first term must be positive, but it is just the reciprocal of the common ratio, so it is positive when the common ratio is, so you get: $x\in (-\infty, -5)\cup (9,\infty)$

For the second part, again you need the common ratio to be less than 1, so do the same procedure as I did for the first one, clearing the denominator, squaring, and solving.

Thanks very much!!!

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Convergent Series

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