Question

\(\sum_{0}^{\infty} 8^{-n}\left(x^{2}-1\right)^{n}\) Method: Let \(y=x^{2}-1\). The power series \(\sum_{0}^{\infty} 8^{-n} y^{n}\) converges for \(|y|<8\), so the original series converges for \(\left|x^{2}-1\right|<8\), which means \(|x|<3\)

Short answer

The series converges for \( \lvert x \rvert < 3 \).

Step-by-step solution

Understand the given series

Given series is \ \(\sum_{0}^{\infty} 8^{-n}\left(x^{2}-1\right)^{n}\)

Let y = x^2 - 1

Substitute \( y = x^2 - 1 \). The series then becomes \ \( \sum_{0}^{\infty} 8^{-n} y^{n} \)

Identify the convergence condition

The power series \( \sum_{0}^{\infty} 8^{-n} y^{n} \) converges for \( \lvert y \rvert < 8 \).

Translate the convergence condition back to x

Since \( y = x^2 - 1 \), replace \( y \) in the inequality: \( \lvert x^2 - 1 \rvert < 8 \).

Solve the inequality for x

Solve \( \lvert x^2 - 1 \rvert < 8 \): \ \begin{align*} -8 &< x^2 - 1 < 8 \ -7 &< x^2 < 9 \ \sqrt{-7} &< \lvert x \rvert < \sqrt{9} \ 0 &< \lvert x \rvert < 3 \ \end{align*} Thus, \( \lvert x \rvert < 3 \).

Express the final result

The given series \( \sum_{0}^{\infty} 8^{-n}\left(x^{2}-1\right)^{n}\) converges for \( \lvert x \rvert < 3 \).

Key concepts

Series Substitution

A common technique in calculus is series substitution. This method simplifies complex series, making them easier to analyze and solve.
In this problem, we start with the given series: \(\sum_{0}^{\infty} 8^{-n}\left(x^{2}-1\right)^{n}\)\
We simplify the expression by introducing a new variable: we can let \(y=x^{2}-1\).
This transforms the series into: \ \(\sum_{0}^{\infty} 8^{-n} y^{n}\)\.By making this substitution, we shift our focus from the original, more complicated series to a simpler power series. This allows us to utilize known results about power series and directly apply them to our problem.

Convergence Condition

Next, we need to determine the convergence condition of the new series. The given series is transformed into a power series: \(\sum_{0}^{\infty} 8^{-n} y^{n}\).
We know that a power series \(\sum_{n=0}^{\infty} a_n y^n\) converges when the absolute value of \(y\) is less than the radius of convergence.
Here, the series converges for \(\lvert y \rvert < 8\).
Now, we need to translate this back to our original variable \(x\). \(y\) was previously defined as \(x^{2}-1\).
We substitute \(y\) back to get: \(\left\| x^2 - 1 \right\| < 8\)\.

Inequalities

Finally, we solve the inequality for \(x\). The transformed inequality is: \(\lvert x^2 - 1 \rvert < 8\).
To solve this inequality, we start by breaking it into two separate inequalities: \(-8 < x^2 - 1 < 8\).
Now we solve each part:

• Add 1 to all parts: \(-8 + 1 < x^2 < 8 + 1\) yields \(-7 < x^2 < 9\)

Then we take the square root of the positive part (since \(x^2\) cannot be negative):

• \(0 < \lvert x \rvert < \sqrt{9}\)
• Thus, we conclude \(\lvert x \rvert < 3\).

This tells us that the original series will converge as long as \(x\) is within this interval.