Find the interval of convergence for the given power series.

The series is convergent on the interval:

Ask by Schwartz Lang. in the United States

Dec 14,2024

Question

Find the interval of convergence for the given power series. $$ \sum_{n=1}^{\infty} \frac{(x-7)^{n}}{n(-4)^{n}} $$ The series is convergent on the interval:

UpStudy AI · Accepted Answer

To find the interval of convergence for the power series

$$
\sum_{n=1}^{\infty} \frac{(x-7)^{n}}{n(-4)^{n}},
$$

we can use the ratio test. The general term of the series is

$$
a_n = \frac{(x-7)^{n}}{n(-4)^{n}}.
$$

We will apply the ratio test, which involves finding the limit of the absolute value of the ratio of consecutive terms:

$$
\left| \frac{a_{n+1}}{a_n} \right|.
$$

Calculating $$a_{n+1}$$:

$$
a_{n+1} = \frac{(x-7)^{n+1}}{(n+1)(-4)^{n+1}} = \frac{(x-7)^{n+1}}{(n+1)(-4)^{n} \cdot (-4)}.
$$

Now, we compute the ratio:

$$
\frac{a_{n+1}}{a_n} = \frac{(x-7)^{n+1}}{(n+1)(-4)^{n+1}} \cdot \frac{n(-4)^{n}}{(x-7)^{n}} = \frac{(x-7) \cdot n}{(n+1)(-4)}.
$$

Taking the absolute value:

$$
\left| \frac{a_{n+1}}{a_n} \right| = \left| \frac{(x-7) \cdot n}{(n+1)(-4)} \right| = \frac{|x-7| \cdot n}{(n+1) \cdot 4}.
$$

Now, we take the limit as $$n$$ approaches infinity:

$$
\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \frac{|x-7| \cdot n}{(n+1) \cdot 4} = \lim_{n \to \infty} \frac{|x-7| \cdot n}{4n + 4} = \lim_{n \to \infty} \frac{|x-7|}{4} \cdot \frac{n}{n + 1} = \frac{|x-7|}{4}.
$$

For the series to converge, we require:

$$
\frac{|x-7|}{4} < 1.
$$

Multiplying both sides by 4 gives:

$$
|x-7| < 4.
$$

This inequality can be rewritten as:

$$
-4 < x - 7 < 4.
$$

Adding 7 to all parts of the inequality results in:

$$
3 < x < 11.
$$

Thus, the interval of convergence is $$ (3, 11) $$.

Next, we need to check the endpoints $$x = 3$$ and $$x = 11$$ to determine if they are included in the interval of convergence.

1. **At $$x = 3$$**:

$$
\sum_{n=1}^{\infty} \frac{(3-7)^{n}}{n(-4)^{n}} = \sum_{n=1}^{\infty} \frac{(-4)^{n}}{n(-4)^{n}} = \sum_{n=1}^{\infty} \frac{1}{n}.
$$

This series diverges (harmonic series).

2. **At $$x = 11$$**:

$$
\sum_{n=1}^{\infty} \frac{(11-7)^{n}}{n(-4)^{n}} = \sum_{n=1}^{\infty} \frac{4^{n}}{n(-4)^{n}} = \sum_{n=1}^{\infty} \frac{1}{n} = \sum_{n=1}^{\infty} \frac{1}{n}.
$$

This series also diverges.

Since both endpoints do not converge, the interval of convergence is:

$$
\boxed{(3, 11)}.
$$
The interval of convergence is $$ (3, 11) $$.

Find the interval of convergence for the given power series.

The series is convergent on the interval:

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