I had to show that the sum of the infinite series 1 - x + x^3 - x^4... is equal to 1/(1 + x + x^2) and state the values of x for which this is valid. I found the first part with no difficulties, but I’m unsure how to find the values for x. The answer states that mod x is less than 1, but I have no idea how they cam cup with that answer. Any help would be appreciated.

I had to show that the sum of the infinite series 1 - x + x^3 - x^4...... is equal to 1/(1 + x + x^2) and state the values of x for which this is valid. I found the first part with no difficulties, but I’m unsure how to find the values for x. The answer states that mod x is less than 1, but I have no idea how they cam cup with that answer. Any help would be appreciated.

I assume you accidentally missed the x^2 term?

Is this the question?

1-x+x^2-x^3+...=\sum_{n=0}^\infty (-x)^n = \frac{1}{1+x+x^2}


The answer states that mod x is less than 1

Mod x? "Mod" to me mean modulo division, which doesn't make a lot of sense in this context. Are you referring to what I would call "absolute value"? (i.e. "mod x" means |x| )?

I assume you accidentally missed the x^2 term?

Is this the question?

1-x+x^2-x^3+...=\sum_{n=0}^\infty (-x)^n = \frac{1}{1+x+x^2}



Mod x? "Mod" to me mean modulo division, which doesn't make a lot of sense in this context. Are you referring to what I would call "absolute value"? (i.e. "mod x" means |x| )?

Hi. Yes sorry, by mod I mean absolute value, but there isn’t an x^2 missing.
First I had to find the sum of the geometric series (1 -x ) + ( x^3 - x^4 ) + ( x^6 - x^7 ) +.. . ( 3^3n - 3^3n-1 ) which I got correctly as [( 1 - x ) ( 1 - x^3n+3)]/( 1 - x^3 )
Then, of this series 1 - x + x^3 - x^4 + x^6... I found the sum to infinity to be ( 1 - x )/ ( 1 - x^3 ) which reduces to 1/ 1 + x + x^2. But I’m stuck when it asked for the values of x for which this is valid. The answer says [x] is less than one, but I can’t understand why it can’t be greater than one. Thanks for your input: really appreciated.

Hi. Yes sorry, by mod I mean absolute value, but there isn’t an x^2 missing.
First I had to find the sum of the geometric series (1 -x ) + ( x^3 - x^4 ) + ( x^6 - x^7 ) + ..... .....( 3^3n - 3^3n-1 ) which I got correctly as [( 1 - x ) ( 1 - x^3n+3)]/( 1 - x^3 )
Then, of this series 1 - x + x^3 - x^4 + x^6...... I found the sum to infinity to be ( 1 - x )/ ( 1 - x^3 ) which reduces to 1/ 1 + x + x^2. But I’m stuck when it asked for the values of x for which this is valid. The answer says [x] is less than one, but I can’t understand why it can’t be greater than one. Thanks for your input: really appreciated.

Since there are an infinite number of terms with ever increasing exponent, the only way the series can possibly converge is for each subsequent term to be smaller than the one before (in other words, the limit of term n as n approaches infinity has to go to zero). Otherwise, each subsequent term is bigger than the last, and the sum grows toward positive or negative infinity ad nauseum.

Since the coefficient of each term is the same (1 in this case -- as opposed to, say, a Taylor series where the coefficient has a 1/n! In it, which can more than compensate for an ever increasing exponent), the only way that the terms will approach zero for large n is for the argument of the exponent (x in this case) to be less than one in magnitude.

\lim_{n\to\infty}x^n = 0 if x<1.

\lim_{n\to\infty}x^n = 1 if x=1.

\lim_{n\to\infty}x^n = \infty if x>1.

The second and third cases (x=1 or x>1) clearly result in non-convergent series.

Likewise, since there are an infinite number of terms,

Since there are an infinite number of terms with ever increasing exponent, the only way the series can possibly converge is for each subsequent term to be smaller than the one before (in other words, the limit of term n as n approaches infinity has to go to zero). Otherwise, each subsequent term is bigger than the last, and the sum grows toward positive or negative infinity ad nauseum.

Since the coefficient of each term is the same (1 in this case -- as opposed to, say, a Taylor series where the coefficient has a 1/n! in it, which can more than compensate for an ever increasing exponent), the only way that the terms will approach zero for large n is for the argument of the exponent (x in this case) to be less than one in magnitude.

\lim_{n\to\infty}x^n = 0 if x<1.

\lim_{n\to\infty}x^n = 1 if x=1.

\lim_{n\to\infty}x^n = \infty if x>1.

The second and third cases (x=1 or x>1) clearly result in non-convergent series.

Likewise, since there are an infinite number of terms,

Really helpful. Thanks so much.