Test the convergence of the series \(\sum_{n=1}^{∞} \left ( \frac{n^n}{n!} \right )\)

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Solution:

Given series is

\(\sum_{n=1}^{∞} \left ( \frac{n^n}{n!} \right )\)

Comparing it with \(\sum_{n=1}^{∞} u_n\), then

\(u_n = \frac{n^n}{n!}\)

Here

\( = \lim_{n \to ∞} \left ( \frac{u_{n+1}}{u_n} \right )\)

\( = \lim_{n \to ∞} \left ( \frac{(n+1)^{n+1}}{(n+1)!} \times \frac{n!}{n^n} \right ) \)

\( = \lim_{n \to ∞} \left ( \frac{(n+1)(n+1)^n}{(n+1)n!} \times \frac{n!}{n^n} \right )  \)

\( = \lim_{n \to ∞} \left ( \frac{n^n(1+\frac{1}{n})^n}{(n+1)n!} \times \frac{1}{n^n} \right ) \)

\( = \lim_{n \to ∞} \left ( 1 + \frac{1}{n} \right )^n \)

= e > 1

This means the given series is divergent by D’Alembert ratio test.

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