I need to write the power series: ∑n=1∞1(x−3)2n−1−1(x−2)2n−1 I need to write it as a rational function. But not sure how

In general ∑n=1∞1u2n−1=uu2−1. Your expression becomes x−3(x−3)2−1−x−2(x−2)2−1=(x−3)((x−2)2−1)−(x−2)((x−3)2−1)((x−3)2−1)((x−2)2−1) You could simplify it. Note that |x-3|&gt;1 and |x-2|&gt;1 required.

This series has the form ∑n=1∞1u2n−1−∑n=1∞1v2n−1=1u∑n=0∞1(u2)n−1v∑n=0∞1(v2)n.

In general ∑n=1∞1u2n−1=uu2−1. Your expression becomes x−3(x−3)2−1−x−2(x−2)2−1=(x−3)((x−2)2−1)−(x−2)((x−3)2−1)((x−3)2−1)((x−2)2−1) You could simplify it. Note that |x-3|&gt;1 and |x-2|&gt;1 required.

This series has the form ∑n=1∞1u2n−1−∑n=1∞1v2n−1=1u∑n=0∞1(u2)n−1v∑n=0∞1(v2)n.

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High School
Algebra
Algebra II
Rational functions

Loretta Cannon

Answered question

2022-02-18

I need to write the power series:

\sum_{n = 1}^{\infty} \frac{1}{{(x - 3)}^{2 n - 1}} - \frac{1}{{(x - 2)}^{2 n - 1}}

I need to write it as a rational function. But not sure how

Answer & Explanation

par5o4nr4z

Beginner2022-02-19Added 5 answers

In general

\sum_{n = 1}^{\infty} \frac{1}{u^{2 n - 1}} = \frac{u}{u^{2} - 1}

. Your expression becomes

\frac{x - 3}{{(x - 3)}^{2} - 1} - \frac{x - 2}{{(x - 2)}^{2} - 1} = \frac{(x - 3) ({(x - 2)}^{2} - 1) - (x - 2) ({(x - 3)}^{2} - 1)}{({(x - 3)}^{2} - 1) ({(x - 2)}^{2} - 1)}

You could simplify it. Note that |x-3|>1 and |x-2|>1 required.

Nathan Kent

Beginner2022-02-20Added 8 answers

This series has the form

\sum_{n = 1}^{\infty} \frac{1}{u^{2 n - 1}} - \sum_{n = 1}^{\infty} \frac{1}{v^{2 n - 1}} = \frac{1}{u} \sum_{n = 0}^{\infty} \frac{1}{{(u^{2})}^{n}} - \frac{1}{v} \sum_{n = 0}^{\infty} \frac{1}{{(v^{2})}^{n}}

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