Question

Use Exercise 3 to explain why the ratio test will be inconclusive for every series $\sum _{k=1}^{\infty }{a}_k$in which $a_k$is a polynomial.

Short answer

If $a_k$is a polynomial then localid="1649104353595" $\underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)}$will be $1$and the ratio test is inconclusive for series when$\underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)}=1.$

Step-by-step solution

Step 1. Given information. 

If $p\left(x\right)$is a polynomial then role="math" localid="1649104261634" $\underset{x\to \infty }{\lim }\frac{p(x+1)}{p\left(x\right)}=1.$

If$a_k$is a polynomial then the ratio test will be inconclusive for every series role="math" localid="1649104105405" $\sum _{k=1}^{\infty }{a}_k.$

Step 2. Verification.

Consider $a_k=p\left(x\right)$

role="math" localid="1649104033456" $$\begin{gathered} a_{k+1}=p(k+1) \\ \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=\underset{k\to \infty }{\lim }\frac{p(k+1)}{p\left(k\right)} \end{gathered}$$

as $\underset{k\to \infty }{\lim }\frac{p(k+1)}{p\left(k\right)}=1$

so role="math" localid="1649104148331" $$\begin{gathered} \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=1 \\ L=1 \end{gathered}$$

According to the ratio test, if $\sum _{k=1}^{\infty }{a}_k$is a series with positive terms and $L=\underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=1$then then the ratio test will be inconclusive for series.