Chapter 36: Chapter 36

Problem 1137

Test the series: \(1+2 ! / 2^{2}+3 ! / 3^{3}+4 ! / 4^{4}+\ldots \ldots\) by means of the ratio test. If this test fails, use another test.

Short Answer

Expert verified

Both the ratio test and the root test were inconclusive for determining the convergence or divergence of the given series \(\sum_{n=1}^{\infty} \frac{n!}{n^n}\). Further analysis using another test or comparison with known convergent or divergent series would be needed to establish convergence or divergence for this series.

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Step by step solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Calculate the ratio of consecutive terms

We are given the series: \(\sum_{n=1}^{\infty} \frac{n!}{n^n}\) Let \(a_n = \frac{n!}{n^n}\), the general term of the series. Now calculate the ratio of consecutive terms, which is \(\frac{a_{n+1}}{a_n}\): \[\frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}\]

Step 2: Simplify the ratio

Multiplying the ratio by the reciprocal of the denominator term, we get: \[\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \cdot \frac{n^n}{n!}\] Simplify the expression: \[\frac{a_{n+1}}{a_n} = \frac{n^n (n+1)!}{n! (n+1)^{n+1}}\]

Step 3: Cancel out terms

Notice the cancellation of factorial terms: \[\frac{a_{n+1}}{a_n} = \frac{n^n (n+1) \cdot n!}{n! (n+1)^{n+1}}\] Now, \(n!\) cancels out: \[\frac{a_{n+1}}{a_n} = \frac{n^n (n+1)}{(n+1)^{n+1}}\]

Step 4: Take the limit as n approaches infinity

Now take the limit as n approaches infinity: \[\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{n^n (n+1)}{(n+1)^{n+1}}\] Divide both numerator and denominator by \(n^n\): \[\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{(n+1)}{(\frac{n+1}{n})^{n+1}}\] Now, apply the limit: \[\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{\infty}{\infty}\] Since the limit is of indeterminate form, the ratio test fails.

Step 5: Apply the Root Test

Since the ratio test failed, let's try applying the root test. The root test states that if the limit of the nth root of the general term exists and is less than 1, the series converges, otherwise, it diverges. Let's find the limit: \[\lim_{n \to \infty} \sqrt[n]{|a_n|} = \lim_{n \to \infty} \sqrt[n]{\frac{n!}{n^n}}\] We can rewrite the expression as: \[\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}}\] Now, apply the limit to the exponent: \[\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}} = \left(\frac{n!}{n^n}\right)^{0}\] Since anything raised to the power of 0 is equal to 1: \[\lim_{n \to \infty} \left(\frac{n!}{n^n}\right)^{\frac{1}{n}} = 1\] Since the limit is equal to 1, the root test is inconclusive. In conclusion, both the ratio test and the root test were inconclusive for determining the convergence or divergence of the series \(\sum_{n=1}^{\infty} \frac{n!}{n^n}\). Further analysis using another test or comparison with known convergent or divergent series would be needed to establish convergence or divergence for this series.

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Chapter 36: Chapter 36

Test the series: \(1+2 ! / 2^{2}+3 ! / 3^{3}+4 ! / 4^{4}+\ldots \ldots\) by means of the ratio test. If this test fails, use another test.

Short Answer

Step by step solution

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Step 1: Calculate the ratio of consecutive terms

Step 2: Simplify the ratio

Step 3: Cancel out terms

Step 4: Take the limit as n approaches infinity

Step 5: Apply the Root Test

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