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

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To determine the values of $$ x $$ for which the series $$ \sum_{k=1}^{\infty}(4 x-1)^{k} $$ converges, we can recognize that this is a geometric series. A geometric series converges if the absolute value of the common ratio is less than 1.

In this case, the common ratio $$ r $$ is given by:

$$
r = 4x - 1
$$

The series converges when:

$$
|4x - 1| < 1
$$

Now, we can solve this inequality step by step.

1. **Set up the inequality**:
   $$
-1 < 4x - 1 < 1
   $$

2. **Break it into two parts**:
   - For the left part:
   $$
   -1 < 4x - 1
   $$
   Adding 1 to both sides:
   $$
   0 < 4x
   $$
   Dividing by 4:
   $$
   0 < x
   $$

- For the right part:
   $$
   4x - 1 < 1
   $$
   Adding 1 to both sides:
   $$
   4x < 2
   $$
   Dividing by 4:
   $$
   x < \frac{1}{2}
   $$

3. **Combine the results**:
   From the two parts, we have:
   $$
   0 < x < \frac{1}{2}
   $$

Thus, the series $$ \sum_{k=1}^{\infty}(4 x-1)^{k} $$ converges for values of $$ x $$ in the interval:

$$
\boxed{(0, \frac{1}{2})}
$$
The series $$ \sum_{k=1}^{\infty}(4x-1)^{k} $$ converges when $$ 0 < x < \frac{1}{2} $$.

4.2 For which values of will

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