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Activity 8.4.2

Avg difficulty 2.5

Consider the infinite Taylor series given by T(x)=∑k=1∞1k⋅2k(x−1)k=11⋅2(x−1)+12⋅4(x−1)2+13⋅8(x−1)3+⋯+1n⋅2n(x−1)n+⋯\begin{aligned} T(x) &= \sum_{k = 1}^{\infty} \frac{1}{k \cdot 2^k}(x-1)^k \\ &= \frac{1}{1 \cdot 2}(x-1) + \frac{1}{2 \cdot 4}(x-1)^2 + \frac{1}{3 \cdot 8}(x-1)^3 + \cdots + \frac{1}{n \cdot 2^n}(x-1)^n + \cdots \end{aligned}

  1. As described in the statement of the Ratio Test, let rn(x)r_n(x) be the ratio of the (n+1)st(n+1)^{\text{st}} term of T(x)T(x) to the nthn^{\text{th}} term of T(x)T(x). Find the simplest formula that you can for rn(x)r_n(x).

  2. Let r(x)=lim⁡n→∞rn(x)r(x) = \lim_{n \to \infty} r_n(x). Evaluate this limit to find the simplest formula you can for r(x)r(x).

  3. For what values of xx is ∣r(x)∣<1|r(x)| \lt 1? What is the open interval of convergence for T(x)T(x)?

  4. Let T10(x)T_{10}(x) be the sum of the first 1010 terms of T(x)T(x), and let f(x)=ln⁡(2)−ln⁡(3−x)f(x) = \ln(2) - \ln(3-x). Plot f(x)f(x) and T10(x)T_{10}(x) on the same coordinate axes in a window centered around x=1x=1. What do you notice? What does this suggest about the series T(x)T(x)?

Boelkins, Active Calculus, §8.4 Taylor series, Activity 8.4.2

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Adapted from Active Calculus © 2012–2025 Matthew Boelkins. Licensed under CC BY-SA 4.0. Source: Active Calculus. Imported at revision dc86ba3.