Q. 2 - Calculus

Question

Construct examples of the thing(s) described in the following. Try to find examples that are different than any in the reading.

(a) A series containing factorials on which the ratio test will be effective in determining convergence or divergence.

(b) A series containing factorials on which the ratio test will be ineffective in determining convergence or divergence.

(c) A series on which the root test will be effective in determining convergence or divergence.

Step-by-Step Solution

Answer

(a) $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }\frac{5^k}{k!}$

(b) $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{2!}{k^2}.$

(c) $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{3^k}{k^5}.$

Part (a) Step 1. Given information.

Consider the following given information.

(a) a factorial, where ratio test can determine whether series is convergence or divergence. 

(b) a factorial series on which the ratio test will be ineffective in determining convergence or divergence. 

(c) A series on which the root test will be effective in determining convergence or divergence. 

Part (a) Step 2. Explanation.

Consider a series $\sum _{k=0}^{\infty }{a}_k=\sum _{k=0}^{\infty }\frac{5^k}{k!}$ and determine the value of $\rho =\underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}.$

$\begin{array}{rcl}\rho & =& \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}\\ & =& \underset{k\to \infty }{\lim }\frac{\frac{5^{k+1}}{(k+1)!}}{\frac{5^k}{k!}} \\ & =& \underset{k\to \infty }{\lim }\frac{5^k·5^1·k!}{5^k·k!·(k+1)}\\ & =& \underset{k\to \infty }{\lim }\frac{5}{k+1}\\ & =& 0\end{array}$

Here $\rho <1,$so the series converges according to the ratio test.

Part (b) Step 1. The explanation for the statement.

Consider a series $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{2!}{k^2}$ and determine the value of $\rho =\underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}.$

$\begin{array}{rcl}\rho & =& \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}\\ & =& \underset{k\to \infty }{\lim }\frac{\frac{2!}{{(k+1)}^2}}{\frac{2!}{k^2}} \\ & =& \underset{k\to \infty }{\lim }\frac{2!·k^2}{2!·{(k+1)}^2}\\ & =& \underset{k\to \infty }{\lim }\frac{k^2}{{(k+1)}^2}\\ & =& 1\end{array}$

Here $\rho =1$so the ratio test is inconclusive.

Part (c) Step 1. The explanation for the statement.

Consider a series $\sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{3^k}{k^5}$ and determine the value of $\rho =\underset{k\to \infty }{\lim }{\left(a_k\right)}^{\frac{1}{k}}$.

$\begin{array}{rcl}\rho & =& \underset{k\to \infty }{\lim }{\left(a_k\right)}^{\frac{1}{k}}\\ & =& \underset{k\to \infty }{\lim }{\left(\frac{3^k}{k^5}\right)}^{\frac{1}{k}}\\ & =& \underset{k\to \infty }{\lim }\left(\frac{3^{{\displaystyle \frac{k}{k}}}}{k^{{\displaystyle \frac{5}{k}}}}\right)\\ & =& 3\end{array}$

Here $\rho >1$so the series converges according to the root test.