Math Snap

STEP 1

1. The function $f(x) = \frac{x^{2}}{(1+7x)^{2}}$ is expressed as a power series.
2. The power series representation is given by $f(x) = \sum_{n=0}^{\infty} \left(7(n+1)(-7)^{n}\right)x^{n+2}$.
3. We need to determine the interval of convergence for this power series.

STEP 2

1. Identify the form of the power series.
2. Use the ratio test to find the interval of convergence.
3. Determine the open interval of convergence.

STEP 3

Identify the form of the power series. The series is given by:
$$f(x) = \sum_{n=0}^{\infty} \left(7(n+1)(-7)^{n}\right)x^{n+2}$$ This series is centered at 0 and involves powers of $x$.

STEP 4

Use the ratio test to find the interval of convergence. The ratio test involves finding:
$$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$$ where $a_n = 7(n+1)(-7)^n x^{n+2}$.

STEP 5

Calculate $a_{n+1}$ and $a_n$:
$$a_n = 7(n+1)(-7)^n x^{n+2}$$ $$a_{n+1} = 7(n+2)(-7)^{n+1} x^{n+3}$$ Now, compute the ratio:
$$\frac{a_{n+1}}{a_n} = \frac{7(n+2)(-7)^{n+1} x^{n+3}}{7(n+1)(-7)^n x^{n+2}}$$ Simplify the expression:
$$= \frac{(n+2)(-7)x}{(n+1)}$$

STEP 6

Apply the limit in the ratio test:
$$\lim_{n \to \infty} \left| \frac{(n+2)(-7)x}{(n+1)} \right| = \lim_{n \to \infty} \left| \frac{(-7)x(n+2)}{(n+1)} \right|$$ As $n \to \infty$, the dominant terms are $n$, so:
$$\lim_{n \to \infty} \left| \frac{(-7)x(n+2)}{(n+1)} \right| = \left| -7x \right|$$ The ratio test tells us that the series converges when:
$$\left| -7x \right| < 1$$

Solve the inequality $\left| -7x \right| < 1$:
$$\left| 7x \right| < 1$$ $$-1 < 7x < 1$$ Divide through by 7:
$$-\frac{1}{7} < x < \frac{1}{7}$$ This gives us the open interval of convergence:
$$\left( -\frac{1}{7}, \frac{1}{7} \right)$$ The open interval of convergence is $\boxed{\left( -\frac{1}{7}, \frac{1}{7} \right)}$.