Question

Show that the series:$\ln \left(2\right)+\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k-1}}{k{2}^k}{\left(x-2\right)}^k$

from Example 3 diverges when x = 0 and converges conditionally when x = 4.

Short answer

$$\begin{gathered} \mathrm{Series}\mathrm{diverges}\mathrm{if}\mathrm{x}=0\mathrm{since}\mathrm{the}\mathrm{harmonic}\mathrm{series}\mathrm{diverges}. \\ \mathrm{Series}\mathrm{converges}\mathrm{conditionally}\mathrm{if}\mathrm{x}=4. \end{gathered}$$

Step-by-step solution

Step 1. Given information is:

$\ln \left(2\right)+\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k-1}}{k{2}^k}{\left(x-2\right)}^k$

Step 2. Check Divergence

$$\begin{gathered} \mathrm{Assume}\mathrm{that}\mathrm{x}=0; \\ \mathrm{f}(\mathrm{x}=0)=\ln \left(2\right)+\sum _{\mathrm{k}=1}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}-1}}{\mathrm{k}{2}^{\mathrm{k}}}{\left(0-2\right)}^{\mathrm{k}} \\ \mathrm{f}(\mathrm{x}=0)=\ln \left(2\right)-\sum _{\mathrm{k}=1}^{\infty }\frac{1}{\mathrm{k}} \\ \mathrm{Therefore},\mathrm{series}\mathrm{diverges}\mathrm{if}\mathrm{x}=0\mathrm{since}\mathrm{the}\mathrm{harmonic}\mathrm{series}\mathrm{diverges}. \end{gathered}$$

Step 3. Check Convergence

$$\begin{gathered} \mathrm{Assume}\mathrm{that}\mathrm{x}=4; \\ \mathrm{f}(\mathrm{x}=4)=\ln \left(2\right)+\sum _{\mathrm{k}=1}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}-1}}{\mathrm{k}{2}^{\mathrm{k}}}{\left(4-2\right)}^{\mathrm{k}} \\ \mathrm{f}(\mathrm{x}=4)=\ln \left(2\right)+\sum _{\mathrm{k}=1}^{\infty }\frac{{\left(-1\right)}^{\mathrm{k}-1}}{\mathrm{k}{2}^{\mathrm{k}}}{\left(2\right)}^{\mathrm{k}} \\ \mathrm{Therefore},\mathrm{series}\mathrm{converges}\mathrm{conditionally}\mathrm{if}\mathrm{x}=4. \end{gathered}$$