Calculus Examples

$\sum_{n = 0}^{\infty} \frac{{(- 2)}^{n}}{n}$

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 $a_{n}$ .

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

Step 3

Simplify.

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Step 3.1

Move the exponent into the absolute value.

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

Step 3.2

Apply the product rule to $\frac{{(- 2)}^{n}}{n}$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Cancel the common factor of $n$ .

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Step 3.3.2.1

Cancel the common factor.

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

Step 3.3.2.2

Rewrite the expression.

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

Step 3.4

Evaluate the exponent.

$L = lim_{n \to \infty} | \frac{- 2}{n^{\frac{1}{n}}} |$

Step 4

Evaluate the limit.

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Step 4.1

Evaluate the limit.

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Step 4.1.1

Move the limit inside the absolute value signs.

$L = | lim_{n \to \infty} \frac{- 2}{n^{\frac{1}{n}}} |$

Step 4.1.2

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

$L = | - 2 lim_{n \to \infty} \frac{1}{n^{\frac{1}{n}}} |$

Step 4.1.3

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

$L = | - 2 \frac{lim_{n \to \infty} 1}{lim_{n \to \infty} n^{\frac{1}{n}}} |$

Step 4.1.4

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

$L = | - 2 \frac{1}{lim_{n \to \infty} n^{\frac{1}{n}}} |$

Step 4.2

Use the properties of logarithms to simplify the limit.

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Step 4.2.1

Rewrite $n^{\frac{1}{n}}$ as $e^{\ln (n^{\frac{1}{n}})}$ .

$L = | - 2 \frac{1}{lim_{n \to \infty} e^{\ln (n^{\frac{1}{n}})}} |$

Step 4.2.2

Expand $\ln (n^{\frac{1}{n}})$ by moving $\frac{1}{n}$ outside the logarithm.

$L = | - 2 \frac{1}{lim_{n \to \infty} e^{\frac{1}{n} \ln (n)}} |$

Step 4.3

Evaluate the limit.

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Step 4.3.1

Move the limit into the exponent.

$L = | - 2 \frac{1}{e^{lim_{n \to \infty} \frac{1}{n} \ln (n)}} |$

Step 4.3.2

Combine $\frac{1}{n}$ and $\ln (n)$ .

$L = | - 2 \frac{1}{e^{lim_{n \to \infty} \frac{\ln (n)}{n}}} |$

Step 4.4

Apply L'Hospital's rule.

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Step 4.4.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 4.4.1.1

Take the limit of the numerator and the limit of the denominator.

$L = | - 2 \frac{1}{e^{\frac{lim_{n \to \infty} \ln (n)}{lim_{n \to \infty} n}}} |$

Step 4.4.1.2

As log approaches infinity, the value goes to $\infty$ .

$L = | - 2 \frac{1}{e^{\frac{\infty}{lim_{n \to \infty} n}}} |$

Step 4.4.1.3

The limit at infinity of a polynomial whose leading coefficient is positive is infinity.

$L = | - 2 \frac{1}{e^{\frac{\infty}{\infty}}} |$

Step 4.4.2

Since $\frac{\infty}{\infty}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{n \to \infty} \frac{\ln (n)}{n} = lim_{n \to \infty} \frac{\frac{d}{d n} [\ln (n)]}{\frac{d}{d n} [n]}$

Step 4.4.3

Find the derivative of the numerator and denominator.

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Step 4.4.3.1

Differentiate the numerator and denominator.

$L = | - 2 \frac{1}{e^{lim_{n \to \infty} \frac{\frac{d}{d n} [\ln (n)]}{\frac{d}{d n} [n]}}} |$

Step 4.4.3.2

The derivative of $\ln (n)$ with respect to $n$ is $\frac{1}{n}$ .

$L = | - 2 \frac{1}{e^{lim_{n \to \infty} \frac{\frac{1}{n}}{\frac{d}{d n} [n]}}} |$

Step 4.4.3.3

Differentiate using the Power Rule which states that $\frac{d}{d n} [n^{n}]$ is $n \cdot n^{n - 1}$ where $n = 1$ .

$L = | - 2 \frac{1}{e^{lim_{n \to \infty} \frac{\frac{1}{n}}{1}}} |$

Step 4.4.4

Multiply the numerator by the reciprocal of the denominator.

$L = | - 2 \frac{1}{e^{lim_{n \to \infty} \frac{1}{n} \cdot 1}} |$

Step 4.4.5

Multiply $\frac{1}{n}$ by $1$ .

$L = | - 2 \frac{1}{e^{lim_{n \to \infty} \frac{1}{n}}} |$

Step 4.5

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

$L = | - 2 \frac{1}{e^{0}} |$

Step 4.6

Simplify the answer.

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Step 4.6.1

Anything raised to $0$ is $1$ .

$L = | - 2 (\frac{1}{1}) |$

Step 4.6.2

Cancel the common factor of $1$ .

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Step 4.6.2.1

Cancel the common factor.

$L = | - 2 (\frac{1}{1}) |$

Step 4.6.2.2

Rewrite the expression.

$L = | - 2 \cdot 1 |$

Step 4.6.3

Multiply $- 2$ by $1$ .

$L = | - 2 |$

Step 4.6.4

The absolute value is the distance between a number and zero. The distance between $- 2$ and $0$ is $2$ .

$L = 2$

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 divergent on $[0, \infty)$

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