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Chapter 1: Q4MP (page 1)

Test for convergence: ${鈭�}_{n = 1}^{鈭�} \frac{2^{n}}{n!}$

Short Answer

Expert verified

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

Step by step solution

01

Concept used to prove that the series converges

The ratio test for convergence is expressed as follows:

$l i m_{n 鈫� 鈭�} \frac{| a_{n + 1} |}{| a_{n} |} < 1$

The ratio test for divergence is expressed as follows:

$l i m_{n 鈫� 鈭�} \frac{| a_{n + 1} |}{| a_{n} |} > 1$

02

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

$n^{th}$ term of the series is, $a_{n} = \frac{{(2)}^{n}}{n!}$

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

$|a_{n}| = \frac{{(2)}^{n}}{n!}$ 鈥︹€� (1)

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

$|a_{n + 1}| = \frac{{(2)}^{n + 1}}{(n + 1)!}$ 鈥︹€� (2)

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

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

Apply the limit in the above ratio as follows:

$l i m_{n 鈫� 鈭�} \frac{|a_{n + 1}|}{|a_{n}|} = l i m_{n 鈫� 鈭�} (\frac{2}{n + 1}) l i m_{n 鈫� 鈭�} \frac{|a_{n + 1}|}{|a_{n}|} = l i m_{n 鈫� 鈭�} (\frac{2}{鈭� + 1}) l i m_{n 鈫� 鈭�} \frac{|a_{n + 1}|}{|a_{n}|} = 0$

Thus, the series converges.

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