Q. 14 - Calculus

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 ______ .

Step-by-Step Solution

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

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 $p<0$.