Finding power series for given values of a sum

**Ulysses** · April 29th 2011, 11:40 AM

I have this exercise which I'm not sure how to solve.
It says: Consider the series $\displaystyle\sum_{n=0}^{\infty}x^n$ Does exists any value of x for which the series converges to five? ¿and to 1/3?

Well, I've reasoned that if there exists that value, then it must be inside of the radius of convergence for the series. So I've found the radius of convergence:

$a_n=1$

$R=\displaystyle\lim_{n \to{}\infty}{\left |{\displaystyle\frac{a_n}{a_{n+1}}}\right |}=1$

But now I don't know how to proceed.

**girdav** · April 29th 2011, 11:48 AM

Did you try to compute the sum?

**Ulysses** · April 29th 2011, 11:51 AM

I'll try.

**Plato** · April 29th 2011, 12:09 PM

Originally Posted by Ulysses

I have this exercise which I'm not sure how to solve. It says: Consider the series $\displaystyle\sum_{0}^{\infty}x^n$ Does exists any value of x for which the series converges to five? ¿and to 1/3?

Hint. If the series converges $\sum\limits_{n = 0}^\infty {x^n } = \frac{1}{{1 - x}}$ .

**topsquark** · April 29th 2011, 12:28 PM

Originally Posted by Ulysses

$\displaystyle\sum_{0}^{\infty}x^n$

Just as a side note for notation... I doubt anyone seriously thinks that this sum is over x. However you have not specified that. Here is the code to make this a sum over n from 0 to infinity:
\displaystyle\sum_{n = 0}^{\infty}x^n

-Dan

**chisigma** · April 29th 2011, 12:32 PM

Finding the sum of the 'geometric series'...

... is based on the 'algebraic identity'...

(1)

... from which follows...

(2)

... so that if |x|<1 is...

(3)

Kind regards

$\chi$ $\sigma$

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