Summation problem.

Problem: solve for $x$, such that:$$\sum_{n=0}^{\infty } x^n=\dfrac{4}{11}$$

My first idea was to evaluate the series: $$\sum_{n=0}^{\infty } x^n=\dfrac{1}{1-x}, \text{ for }|x|<1$$ Then solve the equation: $$\dfrac{1}{1-x}=\dfrac{4}{11}$$ Which gave $x=\dfrac{-7}{4}$, incorrect since $|x|<1$. Therefore, there are no solution for $x$.

I wonder if I've got the correct answer, I'm quite new to this.

Solution 1:

Your answer is correct. In fact, you can set bounds on the formula for geometric sum for $|x| <1$.

We have $$ x \in (-1, 1) \Rightarrow -x \in (-1, 1) $$ $$\Rightarrow 1-x \in (0, 2) $$ $$ \Rightarrow \frac{1}{1-x} \in \left( \frac{1}{2}, \infty \right)$$

Clearly $\tfrac{4}{11}$ doesn't belong to this interval.