Question

Test the following series for convergence.

4.$\sum _{n=1}^{\infty }\frac{{(-3)}^n}{n!}$

Short answer

The series$\sum _{n=1}^{\infty }\frac{{(-3)}^n}{n!}$ converges.

Step-by-step solution

Given information

It is given that the series is$\sum _{n=1}^{\infty }\frac{{(-3)}^n}{n!}$

Convergence Test

According to the convergence test for alternating series: An alternating series converges if the absolute value of the terms decreases steadily to zero, that is, if $\left|a_{n+1}\right|\le \left|a_n\right|$, and$\underset{n\to \infty }{\lim }{a}_n=0$.

Apply properties of Convergence Test

Let $a_n=\frac{{(-3)}^n}{n!}$.

Then, $a_{n+1}=\frac{{(-3)}^{n+1}}{(n+1)!}$.

Apply the convergence test as $\left|a_{n+1}\right|\le \left|a_n\right|$.

$\begin{array}{c}\left|\frac{{(-3)}^{n+1}}{(n+1)!}\right|\le \left|\frac{{(-3)}^n}{n!}\right|\\ \left|\frac{(-3){(-3)}^n}{(n+1)(n!)}\right|\le \left|\frac{{(-3)}^n}{n!}\right|\end{array}$

Since the denominator is bigger, therefore the overall term is lesser.

The first property is satisfied.

Also, for $\underset{n\to \infty }{\lim }{a}_n=0$.

$\begin{array}{l}\underset{n\to \infty }{\lim }\frac{{(-3)}^n}{n!}=0\\ \underset{n\to \infty }{\lim }\frac{{(-3)}^{\infty }}{\left(\infty \right)!}\approx 0\end{array}$

Since the denominator is approaching to infinity, therefore the overall term will approach to zero.

The second property is also satisfied.

Therefore, it is proved that the series converges.