Question

Fill in each blank with an inequality involving p: The series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$converges absolutely if ______ , converges conditionally if ______ , and diverges if ______ .

Short answer

On completing the fill in the blanks, we get, "The series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$converges absolutely if $p>1$, converges conditionally if $p\le 1$, and diverges if $p<0$."

Step-by-step solution

Step 1. Given information.

Consider the given question,

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

Step 2. To fill up the first blank.

The series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$converges absolutely when $\sum _{k=1}^{\infty }\left|\frac{{\left(-1\right)}^{k+1}}{k^p}\right|$converges.

The series role="math" localid="1649238151956" $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}=\sum _{k=1}^{\infty }\left|\frac{{\left(-1\right)}^{k+1}}{k^p}\right|$converges when $p>1$.

Therefore, the series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$converges absolutely for $p>1$.

Hence, the first blank is completed by $p>1$.

Step 3. To fill up the second blank.

The series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$converges conditionally when $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$converges but $\sum _{k=1}^{\infty }\left|\frac{{\left(-1\right)}^{k+1}}{k^p}\right|$diverges.

The series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}=\sum _{k=1}^{\infty }\left|\frac{{\left(-1\right)}^{k+1}}{k^p}\right|$diverges when $p\le 1$.

Therefore, the series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$converges conditionally for role="math" localid="1649238338394" $\mathrm{p}\le 1$.

Hence, the second blank is completed by $p\le 1$.

Step 4. To fill up the third blank.

The series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$diverges.

The series $\sum _{k=1}^{\infty }\frac{{\left(-1\right)}^{k+1}}{k^p}$when $p<0$.

Hence, the second blank is completed by$p<0$.