Q. 3 - Calculus

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.

Step-by-Step Solution

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 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

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