4. For which values of $$ x $$ will $$ \sum_{k=1}^{\infty}(4 x-1)^{k} $$ exists.

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To determine for which values of $$ x $$ the series $$ \sum_{k=1}^{\infty}(4x-1)^{k} $$ exists, we need to consider the convergence of the series. The series $$ \sum_{k=1}^{\infty}(4x-1)^{k} $$ is a geometric series with the first term $$ a = 4x - 1 $$ and the common ratio $$ r = 4x - 1 $$. A geometric series converges if and only if the absolute value of the common ratio is less than 1. Therefore, for the series to exist, we need to find the values of $$ x $$ for which $$ |4x - 1| < 1 $$. Let's solve the inequality $$ |4x - 1| < 1 $$ to find the range of values for $$ x $$ that make the series converge. Solve the inequality by following steps: - step0: Solve for $$x$$: $$\left|4x-1 ight|<1$$ - step1: Separate into possible cases: $$\left\{ \begin{array}{l}4x-1<1\4x-1>-1\end{array} ight.$$ - step2: Solve the inequality: $$\left\{ \begin{array}{l}x<\frac{1}{2}\x>0\end{array} ight.$$ - step3: Find the intersection: $$0The series $$ \sum_{k=1}^{\infty}(4x-1)^{k} $$ exists when $$ 0 < x < \frac{1}{2} $$.

For which values of will exists.

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