Determine if Convergent Using Cauchy's Root Test

$\sum_{n = 1}^{\infty} {(\frac{2 n + n^{3}}{5 n^{3} + 1})}^{n}$

Step 1

For an infinite series $\sum_{}^{} a_{n}$ , find the limit $L = lim_{n \to \infty} {| a_{n} |}^{\frac{1}{n}}$ to determine convergence using Cauchy's Root Test.

$L = lim_{n \to \infty} {| a_{n} |}^{\frac{1}{n}}$

Step 2

Substitute for $a_{n}$ .

$L = lim_{n \to \infty} {| {(\frac{2 n + n^{3}}{5 n^{3} + 1})}^{n} |}^{\frac{1}{n}}$

Step 3

Simplify.

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Step 3.1

Move the exponent into the absolute value.

$L = lim_{n \to \infty} | {({(\frac{2 n + n^{3}}{5 n^{3} + 1})}^{n})}^{\frac{1}{n}} |$

Step 3.2

Multiply the exponents in ${({(\frac{2 n + n^{3}}{5 n^{3} + 1})}^{n})}^{\frac{1}{n}}$ .

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Step 3.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$L = lim_{n \to \infty} | {(\frac{2 n + n^{3}}{5 n^{3} + 1})}^{n \frac{1}{n}} |$

Step 3.2.2

Cancel the common factor of $n$ .

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Step 3.2.2.1

Cancel the common factor.

$L = lim_{n \to \infty} | {(\frac{2 n + n^{3}}{5 n^{3} + 1})}^{n \frac{1}{n}} |$

Step 3.2.2.2

Rewrite the expression.

$L = lim_{n \to \infty} | {(\frac{2 n + n^{3}}{5 n^{3} + 1})}^{1} |$

Step 3.3

Simplify.

$L = lim_{n \to \infty} | \frac{2 n + n^{3}}{5 n^{3} + 1} |$

Step 4

Evaluate the limit.

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Step 4.1

Move the limit inside the absolute value signs.

$L = | lim_{n \to \infty} \frac{2 n + n^{3}}{5 n^{3} + 1} |$

Step 4.2

Divide the numerator and denominator by the highest power of $n$ in the denominator, which is $n^{3}$ .

$L = | lim_{n \to \infty} \frac{\frac{2 n}{n^{3}} + \frac{n^{3}}{n^{3}}}{\frac{5 n^{3}}{n^{3}} + \frac{1}{n^{3}}} |$

Step 4.3

Evaluate the limit.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Cancel the common factor of $n$ and $n^{3}$ .

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Step 4.3.1.1.1

Factor $n$ out of $2 n$ .

$L = | lim_{n \to \infty} \frac{\frac{n \cdot 2}{n^{3}} + \frac{n^{3}}{n^{3}}}{\frac{5 n^{3}}{n^{3}} + \frac{1}{n^{3}}} |$

Step 4.3.1.1.2

Cancel the common factors.

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Step 4.3.1.1.2.1

Factor $n$ out of $n^{3}$ .

$L = | lim_{n \to \infty} \frac{\frac{n \cdot 2}{n \cdot n^{2}} + \frac{n^{3}}{n^{3}}}{\frac{5 n^{3}}{n^{3}} + \frac{1}{n^{3}}} |$

Step 4.3.1.1.2.2

Cancel the common factor.

$L = | lim_{n \to \infty} \frac{\frac{n \cdot 2}{n \cdot n^{2}} + \frac{n^{3}}{n^{3}}}{\frac{5 n^{3}}{n^{3}} + \frac{1}{n^{3}}} |$

Step 4.3.1.1.2.3

Rewrite the expression.

$L = | lim_{n \to \infty} \frac{\frac{2}{n^{2}} + \frac{n^{3}}{n^{3}}}{\frac{5 n^{3}}{n^{3}} + \frac{1}{n^{3}}} |$

Step 4.3.1.2

Cancel the common factor of $n^{3}$ .

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Step 4.3.1.2.1

Cancel the common factor.

$L = | lim_{n \to \infty} \frac{\frac{2}{n^{2}} + \frac{n^{3}}{n^{3}}}{\frac{5 n^{3}}{n^{3}} + \frac{1}{n^{3}}} |$

Step 4.3.1.2.2

Rewrite the expression.

$L = | lim_{n \to \infty} \frac{\frac{2}{n^{2}} + 1}{\frac{5 n^{3}}{n^{3}} + \frac{1}{n^{3}}} |$

Step 4.3.2

Cancel the common factor of $n^{3}$ .

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Step 4.3.2.1

Cancel the common factor.

$L = | lim_{n \to \infty} \frac{\frac{2}{n^{2}} + 1}{\frac{5 n^{3}}{n^{3}} + \frac{1}{n^{3}}} |$

Step 4.3.2.2

Divide $5$ by $1$ .

$L = | lim_{n \to \infty} \frac{\frac{2}{n^{2}} + 1}{5 + \frac{1}{n^{3}}} |$

Step 4.3.3

Split the limit using the Limits Quotient Rule on the limit as $n$ approaches $\infty$ .

$L = | \frac{lim_{n \to \infty} \frac{2}{n^{2}} + 1}{lim_{n \to \infty} 5 + \frac{1}{n^{3}}} |$

Step 4.3.4

Split the limit using the Sum of Limits Rule on the limit as $n$ approaches $\infty$ .

$L = | \frac{lim_{n \to \infty} \frac{2}{n^{2}} + lim_{n \to \infty} 1}{lim_{n \to \infty} 5 + \frac{1}{n^{3}}} |$

Step 4.3.5

Move the term $2$ outside of the limit because it is constant with respect to $n$ .

$L = | \frac{2 lim_{n \to \infty} \frac{1}{n^{2}} + lim_{n \to \infty} 1}{lim_{n \to \infty} 5 + \frac{1}{n^{3}}} |$

Step 4.4

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{n^{2}}$ approaches $0$ .

$L = | \frac{2 \cdot 0 + lim_{n \to \infty} 1}{lim_{n \to \infty} 5 + \frac{1}{n^{3}}} |$

Step 4.5

Evaluate the limit.

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Step 4.5.1

Evaluate the limit of $1$ which is constant as $n$ approaches $\infty$ .

$L = | \frac{2 \cdot 0 + 1}{lim_{n \to \infty} 5 + \frac{1}{n^{3}}} |$

Step 4.5.2

Split the limit using the Sum of Limits Rule on the limit as $n$ approaches $\infty$ .

$L = | \frac{2 \cdot 0 + 1}{lim_{n \to \infty} 5 + lim_{n \to \infty} \frac{1}{n^{3}}} |$

Step 4.5.3

Evaluate the limit of $5$ which is constant as $n$ approaches $\infty$ .

$L = | \frac{2 \cdot 0 + 1}{5 + lim_{n \to \infty} \frac{1}{n^{3}}} |$

Step 4.6

Since its numerator approaches a real number while its denominator is unbounded, the fraction $\frac{1}{n^{3}}$ approaches $0$ .

$L = | \frac{2 \cdot 0 + 1}{5 + 0} |$

Step 4.7

Simplify the answer.

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Step 4.7.1

Simplify the numerator.

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Step 4.7.1.1

Multiply $2$ by $0$ .

$L = | \frac{0 + 1}{5 + 0} |$

Step 4.7.1.2

Add $0$ and $1$ .

$L = | \frac{1}{5 + 0} |$

Step 4.7.2

Add $5$ and $0$ .

$L = | \frac{1}{5} |$

Step 4.7.3

$\frac{1}{5}$ is approximately $0.2$ which is positive so remove the absolute value

$L = \frac{1}{5}$

Step 4.8

Divide $1$ by $5$ .

$L = 0.2$

Step 5

If $L < 1$ , the series is absolutely convergent. If $L > 1$ , the series is divergent. If $L = 1$ , the test is inconclusive. In this case, $L < 1$ .

The series is convergent on $[1, \infty)$

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