2.1 Given: $$ x+\frac{1}{2} x^{2}+\frac{1}{4} x^{3}+\frac{1}{8} x^{4}+\ldots $$ For which values of $$ x $$ will the series converge? Show working details.

Question

UpStudy AI · Accepted Answer

To determine the values of $$ x $$ for which the series

$$
S = x + \frac{1}{2} x^{2} + \frac{1}{4} x^{3} + \frac{1}{8} x^{4} + \ldots
$$

converges, we can recognize that this series can be expressed in a more manageable form.

### Step 1: Identify the series type

The series can be rewritten as:

$$
S = \sum_{n=1}^{\infty} \frac{1}{2^{n-1}} x^{n}
$$

This is a power series where the general term is given by:

$$
a_n = \frac{1}{2^{n-1}} x^{n}
$$

### Step 2: Apply the ratio test

To find the radius of convergence, we can use the ratio test. We compute the limit of the absolute value of the ratio of consecutive terms:

$$
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{\frac{1}{2^n} x^{n+1}}{\frac{1}{2^{n-1}} x^{n}} \right| = \left| \frac{1}{2^n} \cdot \frac{1}{\frac{1}{2^{n-1}}} \cdot x \right| = \left| \frac{x}{2} \right|
$$

### Step 3: Set up the convergence condition

According to the ratio test, the series converges if:

$$
\left| \frac{x}{2} \right| < 1
$$

This simplifies to:

$$
|x| < 2
$$

### Step 4: Conclusion

Thus, the series converges for:

$$
-2 < x < 2
$$

### Final Answer

The series converges for values of $$ x $$ in the interval:

$$
(-2, 2)
$$
The series converges for $$ x $$ values between -2 and 2.

2.1 Given:
For which values of will the series converge? Show working details.

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2.1 Given: For which values of will the series converge? Show working details.