Question

Find the radius of convergence, R, of the series. ∞ (−1)n (x − 6)n 4n + 1 n = 0 R =
Find the interval, I, of convergence of the series. (Enter your answer using interval notation.) I =

Answer 1

I'm guessing the series is

$\displaystyle\sum_{n=0}^\infty\frac{(-1)^n(x-6)^n}{4n+1}$

Use the ratio test: we have

$\displaystyle\lim_{n\to\infty}\left|\frac{\frac{(-1)^{n+1}(x-6)^{n+1}}{4n+5}}{\frac{(-1)^n(x-6)^n}{4n+1}}\right|=|x-6|\lim_{n\to\infty}\left|\frac{4n+1}{4n+5}\right|=|x-6|$

The series converges for

$|x-6|<1\implies-1<x-6<1\implies5<x<7$

which has a radius of convergence of 1.

Answer 2

The radius of convergence of $\sum\limits^{\infty}_{n=0} \frac{(-1)^n(x-6)^n}{4n+1}$ is 1, and the interval notation is (5,7)

The radius and interval of convergence

The series is given as:

$\sum\limits^{\infty}_{n=0} \frac{(-1)^n(x-6)^n}{4n+1}$

The ratio test of a series states that:

$L = \lim_{n \to \infty} |\frac{a_{n+1}}{a_n} |$

• If L < 1, then $\sum a_n$ converges
• If L > 1, then $\sum a_n$ diverges
• If L = 1, then $\sum a_n$ is inconclusive

By applying the ratio test considering that the series converges, we have:

$\lim_{n \to \infty} | \frac{(-1)^{n+1}(x-6)^{n+1}}{4(n+1)+1} \div \frac{(-1)^n(x-6)^n}{4n+1}|$

Open the bracket

$\lim_{n \to \infty} | \frac{(-1)^{n+1}(x-6)^{n+1}}{4n+4+1} \div \frac{(-1)^n(x-6)^n}{4n+1}|$

Evaluate the sum

$\lim_{n \to \infty} | \frac{(-1)^{n+1}(x-6)^{n+1}}{4n+5} \div \frac{(-1)^n(x-6)^n}{4n+1}|$

Express as product

$\lim_{n \to \infty} | \frac{(-1)^{n+1}(x-6)^{n+1}}{4n+5} \times \frac{4n+1}{(-1)^n(x-6)^n}|$

Expand the exponents

$\lim_{n \to \infty} | \frac{(-1)^{n} \times (-1)^1 (x-6)^{n} \times (x-6)}{4n+5} \times \frac{4n+1}{(-1)^n(x-6)^n}|$

Cancel out the common factors

$\lim_{n \to \infty} | \frac{(-1)(x-6)(4n+1)}{4n+5} |$

Factor out (-1)(x - 6)

$|x-6|\lim_{n \to \infty} | \frac{(4n+1)}{4n+5} |$

Assume the limit to infinity is 1, the equation becomes

$|x- 6| < 1$ --- condition for convergence series

Split

$-1 < x - 6 < 1$

Add 6 to both sides

$5 < x < 7$

Rewrite as interval notation

(5,7)

Hence, the radius of convergence is 1, and the interval notation is (5,7)

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