Q. 24 - Calculus

Question

Use the alternating series test to determine whether the series in Exercises 24–29 converge or diverge. If a series converges, determine whether it converges absolutely or conditionally.

$\sum _{k=1}^{\mathrm{\infty }} \frac{(-k{)}^k}{k!}$

Step-by-Step Solution

Answer

The series $\sum _{k=1}^{\mathrm{\infty }} \frac{(-k{)}^k}{k!}$diverges

Step 1. Given information

$\sum _{k=1}^{\mathrm{\infty }} \frac{(-k{)}^k}{k!}$

Step 2. Substitute k = k + 1 in a k = ( k ) k k !

$$\begin{gathered} \sum _{k=1}^{\mathrm{\infty }} \frac{(-k{)}^k}{k!}=\sum _{k=1}^{\mathrm{\infty }} \frac{(-1{)}^k(k{)}^k}{k!} \\ \sum _{k=1}^{\mathrm{\infty }} \frac{(-k{)}^k}{k!}=\sum _{k=1}^{\mathrm{\infty }} (-1{)}^k{a}_k \end{gathered}$$

Now,

$a_{k+1}=\frac{(k+1{)}^{k+1}}{(k+1)!}$

Step 3. Adding the limit

$\begin{array}{r}\underset{k\to \mathrm{\infty }}{lim} a_k=\underset{k\to \mathrm{\infty }}{lim} \frac{(k{)}^k}{k!}\\ \ne 0\end{array}$