Find out the answer to "LetS(x)=∑n=1∞4n(x+4)2nnFind the radius of convergence."

Ayden Case

Answered question

2022-02-22

Let

S (x) = \sum_{n = 1}^{\infty} \frac{4^{n} {(x + 4)}^{2 n}}{n}

Find the radius of convergence.

Cicolinif73

Beginner2022-02-23Added 7 answers

Let

S_{N} (x)

denote the partial sums of the series

S (x) = \sum_{n = 1}^{\infty} \frac{4^{n} {(x + 2)}^{2 n}}{n}

. Then, letting

y = {(2 x + 8)}^{2}

, with

| y | < 1

, we have

S_{N} (x) = \sum_{n = 1}^{N} \frac{4^{n} {(x + 2)}^{2 n}}{n}

= \sum_{n = 1}^{N} \frac{y^{n}}{n}

= \sum_{n = 1}^{N} \int_{0}^{y} z^{n - 1} d z

= \int_{0}^{y} \frac{1 - z^{N}}{1 - z} d z

\to - \log | 1 - y |

N \to \infty

(1)

= - \log | 1 - {(2 x + 8)}^{2} |

= - \log | 2 x + 9 | - \log | 2 x + 7 |

where the justification for interchanging the limit with the integral in (1) is provided by the Dominated Convergence Theorems since

| \frac{1 - z^{N}}{1 - z} | \leq ∣ \frac{2}{| 1 - z |}

for

| z | < 1

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