Question

Explain why you must use two convergence tests to show that a series$\sum _{k=1}^{\infty }{a}_{\mathrm{k}}\mathrm{Converges}\mathrm{conditionally}$

Short answer

$\mathrm{For}\mathrm{conditionally}\mathrm{convergent}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }\left|{\mathrm{a}}_{\mathrm{k}}\right|\mathrm{must}\mathrm{diverge}\mathrm{and}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}\mathrm{converges}.$

Step-by-step solution

Step 1. Given

$\sum _{k=1}^{\infty }{a}_{\mathrm{k}}\mathrm{Converges}\mathrm{conditionally}$.

Step 2. Explanation

$$\begin{gathered} \mathrm{The}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}\mathrm{will}\mathrm{be}\mathrm{converge}\mathrm{conditionally}\mathrm{if}\sum _{\mathrm{k}=1}^{\infty }\left|{\mathrm{a}}_{\mathrm{k}}\right|\mathrm{Diverges}. \\ \mathrm{Hence},\mathrm{for}\mathrm{conditionally}\mathrm{convergent}\mathrm{series}\sum _{\mathrm{k}=1}^{\infty }\left|{\mathrm{a}}_{\mathrm{k}}\right|\mathrm{must}\mathrm{diverge}\mathrm{and}\sum _{\mathrm{k}=1}^{\infty }{\mathrm{a}}_{\mathrm{k}}\mathrm{converges}. \\ \mathrm{Hence},\mathrm{for}\mathrm{conditionally}\mathrm{convergent}\mathrm{series}\mathrm{two}\mathrm{test}\mathrm{are}\mathrm{required}. \end{gathered}$$