Chapter 1: Q4MP (page 1)

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

Short Answer

Expert verified

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:

$l i m_{n \to \infty} \frac{| a_{n + 1} |}{| a_{n} |} < 1$

The ratio test for divergence is expressed as follows:

$l i m_{n \to \infty} \frac{| a_{n + 1} |}{| a_{n} |} > 1$

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:

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

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

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

$l i m_{n \to \infty} \frac{|a_{n + 1}|}{|a_{n}|} = l i m_{n \to \infty} (\frac{2}{n + 1}) l i m_{n \to \infty} \frac{|a_{n + 1}|}{|a_{n}|} = l i m_{n \to \infty} (\frac{2}{\infty + 1}) l i m_{n \to \infty} \frac{|a_{n + 1}|}{|a_{n}|} = 0$

Thus, the series converges.

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

Short Answer

Step by step solution

Concept used to prove that the series converges

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

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Test for convergence:∑n=1∞2nn!

Short Answer

Step by step solution

Concept used to prove that the series converges

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

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Electromagnetism

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Study anywhere. Anytime. Across all devices.

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