Question

In Exercises 35–40 use the root test to analyze whether the given series converges or diverges. If the root test is inconclusive, use a different test to analyze the series.

$\sum _{k=1}^{\infty }\frac{k^k}{k!}$

Short answer

The given series diverges.

Step-by-step solution

Step 1. Given Information.

The given series is$\sum _{k=1}^{\infty }\frac{k^k}{k!}.$

Step 2. Determine whether the given series converges or diverges. 

By using the root test, let the general term is $a_k=\frac{k^k}{k!}.$

So,

$$\begin{gathered} \rho =\underset{k\to \infty }{\lim }{\left(a_k\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.} \\ \rho =\underset{k\to \infty }{\lim }{\left(\frac{k^k}{k!}\right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$k$}\right.} \\ \rho =\underset{k\to \infty }{\lim }\left(\frac{k}{{\left(k!\right)}^{{\displaystyle \frac{1}{k}}}}\right) \\ \rho =\infty \end{gathered}$$

Since $\infty >1,$by using the root test the given series diverges.