Question

Explain why the ratio test does not work in determining the convergence or divergence of the series $\sum _{k=1}^{\infty }\frac{k!}{(k+2)!}$. What test would be more effective to analyze this series?

Short answer

Hence proved that the ratio test will be inconclusive because$L=1$and Convergence and Divergence test can be used.

Step-by-step solution

Step 1. Given information.

We are given$\sum _{k=1}^{\infty }\frac{k!}{(k+2)!}$.

Step 2. Ratio Test.

On using Ratio Test,

$$\begin{gathered} a_{k+1}=\frac{(k+1)!}{(k+3)!} \\ \frac{a_{k+1}}{a_k}=\frac{\frac{(k+1)!}{(k+3)!}}{\frac{k!}{(k+2)!}} \\ \frac{a_{k+1}}{a_k}=\frac{(k+1)k!(k+2)!}{(k+3)(k+2)!k!} \\ =\frac{k+1}{k+3} \\ \underset{k\to \infty }{\lim }\frac{a_{k+1}}{a_k}=\underset{k\to \infty }{\lim }\frac{k+1}{k+3} \\ =\underset{k\to \infty }{\lim }\frac{k\left(1+\frac{1}{k}\right)}{k\left(1+\frac{3}{k}\right)} \\ L=1 \\ Therefore,thetestisinconclusive. \end{gathered}$$

Step 3. Convergence and Divergence test.

Now,

$$\begin{gathered} \sum _{k=1}^{\infty }{a}_k=\sum _{k=1}^{\infty }\frac{k!}{(k+2)!}\text{and}\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^2} \\ \text{Clearly,}0\le \frac{k!}{(k+2)!}\le \frac{1}{k^2}\text{.} \\ \text{Now,}\sum _{k=1}^{\infty }{b}_k-\sum _{k=1}^{\infty }\frac{1}{k^2}\text{is of the form}\sum _{k=1}^{\infty }{b}_k-\sum _{k=1}^{\infty }\frac{1}{k^p}\text{which ie a}p\text{eeries.} \\ \text{Here,}p=2>1\text{.} \\ \text{Hence, the series}\sum _{k=1}^{\infty }{b}_k=\sum _{k=1}^{\infty }\frac{1}{k^2}\text{converges.} \end{gathered}$$