Question

Test for convergence:$\mathbf{\sum }_{\mathbf{n}\mathbf{=}\mathbf{1}}^{\mathbf{\infty }}\frac{{\mathbf{2}}^{\mathbf{n}}}{\mathbf{n}\mathbf{!}}$

Short answer

The series$\sum _{n=1}^{\infty }\frac{2^n}{n!}$ converges.

Step-by-step solution

Concept used to prove that the series converges

The ratio test for convergence is expressed as follows:

$\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{n}\mathbf{\to }\mathbf{\infty }}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}\mathbf{<}\mathbf{1}$

The ratio test for divergence is expressed as follows:

$\mathit{l}\mathit{i}{\mathit{m}}_{\mathbf{n}\mathbf{\to }\mathbf{\infty }}\frac{\left|a_{n+1}\right|}{\left|a_n\right|}\mathbf{>}\mathbf{1}$

Calculation to prove that the series  ∑n=1∞2nn!converges

${\text{n}}^{\text{th}}$term of the series is, $a_n=\frac{{\left(2\right)}^n}{n!}$

The magnitude of ${\text{n}}^{\text{th}}$terms is:

$\left|a_n\right|=\frac{{\left(2\right)}^n}{n!}$ …… (1)

The magnitude of$(n+1{)}^{th}$terms is calculated as follows:

$\left|a_{n+1}\right|=\frac{{\left(2\right)}^{n+1}}{(n+1)!}$ …… (2)

Take the ratio of equation (1) by (2) as follows:

$$\begin{gathered} \frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{{\left(2\right)}^{n+1}}{(n+1)!}}{\frac{{\left(2\right)}^n}{n!}} \\ \frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{\frac{2{\left(2\right)}^n}{(n+1)n!}}{\frac{{\left(2\right)}^n}{n!}} \\ \frac{\left|a_{n+1}\right|}{\left|a_n\right|}=\frac{2}{n+1} \end{gathered}$$

Apply the limit in the above ratio as follows:

$$\begin{gathered} li{m}_{n\to \infty }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=li{m}_{n\to \infty }\left(\frac{2}{n+1}\right) \\ li{m}_{n\to \infty }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=li{m}_{n\to \infty }\left(\frac{2}{\infty +1}\right) \\ li{m}_{n\to \infty }\frac{\left|a_{n+1}\right|}{\left|a_n\right|}=0 \end{gathered}$$

Thus, the series converges.