Alternating Series with Error Bound - AP Calculus BC

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Determine whether the series converges or diverges:

$\sum_${n=0}^{\infty }$$(-1)^{n+1}$$\left($\frac{n^4$$}{2n^4$+1}$\right)$

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To determine whether the series converges or diverges, we must use the Alternating Series test, which states that for

$\sum $a_{n}$$ - and $a_${n}=(-1)^${n+1}$b_{n}$$ where for all n - to converge,

must equal zero and must be a decreasing series.

For our series,

because it behaves like

.

The test fails because $$\frac{1}{2}$\neq0$ so we do not need to check the second condition of the test.

The series is divergent.

Determine whether the following series converges or diverges:

$\sum_${n=0}^{\infty }$$(-1)^n$($\frac{5}{n}$)$

Tap to see back →

Given just the harmonic series, we would state that the series diverges. However, we are given the alternating harmonic series. To determine whether this series will converge or diverge, we must use the Alternating Series test.

The test states that for a given series $\sum_${n=0}^{\infty $}$a_{n}$$ where $a_${n}=(-1)^$n$(b_{n}$)$ or $a_${n}=(-1)^${n+1}$(b_{n}$)$ where for all n, if and is a decreasing sequence, then $\sum_${n=0}^{\infty $}$a_{n}$$ is convergent.

First, we must evaluate the limit of as n approaches infinity:

The limit equals zero because the numerator of the fraction equals zero as n approaches infinity.

Next, we must determine if is a decreasing sequence. $\frac{1}{n}$< $\frac{1}{n+1}$ , thus the sequence is decreasing.

Because both parts of the test passed, the series is (absolutely) convergent.

Which of the following series does not converge?

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We can show that the series $\sum_${i=0}^{\infty}$ $$\frac{n!}{n^2$ cos(n)}$$ diverges using the ratio test.

$L = $\lim_{n ->\infty }$ $$\frac{a_{n+1}$$$}{a_{n}$} = $\lim_{n ->\infty }$ $\left[\left($\frac{(n+1)!}{(n+1)^2$ cos (n+1)}$ \div $$\frac{n!}{n^2$ cos (n) }$\right)\right]$

$= $\lim_{n ->\infty}$ $$\frac{n^2$$(n+1)cos(n)}{(n+1)^2cos$(n+1)}$ = $\lim_{n ->\infty}$ $$\frac{n^2cos$(n)}{(n+1)cos(n+1)}$$

will dominate over since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that is much greater than , and thus having in the numerator will make the series diverge by the limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

Determine whether

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges or diverges, and explain why.

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We can use the alternating series test to show that

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges.

We must have $\small \sin $\frac{1}{n}$\geq 0$ for $\small n\geq 1$ in order to use this test. This is easy to see because $\small $\frac{1}{n}$$ is in $\small \small \left[0,$\frac{\pi}{2}\right]$ for all $\small n\geq 1$ (the values of this sequence are $\small 1,1/2,1/3,1/4,...$ ), and sine is always nonzero whenever sine's argument is in $\small \small \left[0,$\frac{\pi}{2}\right]$ .

Now we must show that

1. $\small \small $\lim_{n\to\infty}$ \sin $\frac{1}{n}$=0$

2. $\small \sin $\frac{1}{n}$$ is a decreasing sequence.

The limit

$\small $\lim_{n\to\infty}$$\frac{1}{n}$=0$

implies that

$\small \small \small $\lim_{n\to\infty}$ \sin $\frac{1}{n}$=\sin 0=0$

so the first condition is satisfied.

We can show that $\small \small \small \sin $\frac{1}{n}$$ is decreasing by taking its derivative and showing that it is less than $\small 0$ for $\small n\geq 1$ :

$\small \small \small \small \small $\frac{d}{dn}$\sin $$\frac{1}{n}$=-$\frac{1}{n^2$}$\cos$\frac{1}{n}$$

The derivative is less than $\small 0$ , because $\small $-$\frac{1}{n^2$}$$ is always less than $\small 0$ , and that $\small \cos $\frac{1}{n}$$ is positive for $\small n\geq 1$ , using a similar argument we used to prove that $\small \small \sin $\frac{1}{n}$\geq 0$ for $\small n\geq 1$ . Since the derivative is less than $\small 0$ , $\small \small \small \sin $\frac{1}{n}$$ is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges, by the alternating series test.

For the series: $\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n^n$$}{8^n$}$)$ , determine if the series converge or diverge. If it diverges, choose the best reason.

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The series given is an alternating series.

Write the three rules that are used to satisfy convergence in an alternating series test.

For $\sum_${n=1}^{\infty}$ $(-1)^{n+1}$ x_n= x_1-x_2+x_3-x_4+...$ :

$\textup{1. All the } x_n \textup{ terms are positive.}$

$\textup{2. The terms must be decreasing: } x_n \geq $x_{n+1}$$

$\textup{3. The limit $}\lim_{n \to \infty}$x_n=0.$

$\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n^n$$}{8^n$}$)=\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n}{8}$)^n$$

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since and instead approaches infinity.

The correct answer is:

$\textup{Diverge, $since}\lim_{n \to \infty}$x_n\neq 0.$

Determine whether the series converges or diverges:

$\sum_${n=0}^{\infty }$$(-1)^{n+1}$$\left($\frac{n^4$$}{2n^4$+1}$\right)$

Tap to see back →

To determine whether the series converges or diverges, we must use the Alternating Series test, which states that for

$\sum $a_{n}$$ - and $a_${n}=(-1)^${n+1}$b_{n}$$ where for all n - to converge,

must equal zero and must be a decreasing series.

For our series,

because it behaves like

.

The test fails because $$\frac{1}{2}$\neq0$ so we do not need to check the second condition of the test.

The series is divergent.

For the series: $\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n^n$$}{8^n$}$)$ , determine if the series converge or diverge. If it diverges, choose the best reason.

Tap to see back →

The series given is an alternating series.

Write the three rules that are used to satisfy convergence in an alternating series test.

For $\sum_${n=1}^{\infty}$ $(-1)^{n+1}$ x_n= x_1-x_2+x_3-x_4+...$ :

$\textup{1. All the } x_n \textup{ terms are positive.}$

$\textup{2. The terms must be decreasing: } x_n \geq $x_{n+1}$$

$\textup{3. The limit $}\lim_{n \to \infty}$x_n=0.$

$\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n^n$$}{8^n$}$)=\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n}{8}$)^n$$

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since and instead approaches infinity.

The correct answer is:

$\textup{Diverge, $since}\lim_{n \to \infty}$x_n\neq 0.$

Which of the following series does not converge?

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We can show that the series $\sum_${i=0}^{\infty}$ $$\frac{n!}{n^2$ cos(n)}$$ diverges using the ratio test.

$L = $\lim_{n ->\infty }$ $$\frac{a_{n+1}$$$}{a_{n}$} = $\lim_{n ->\infty }$ $\left[\left($\frac{(n+1)!}{(n+1)^2$ cos (n+1)}$ \div $$\frac{n!}{n^2$ cos (n) }$\right)\right]$

$= $\lim_{n ->\infty}$ $$\frac{n^2$$(n+1)cos(n)}{(n+1)^2cos$(n+1)}$ = $\lim_{n ->\infty}$ $$\frac{n^2cos$(n)}{(n+1)cos(n+1)}$$

will dominate over since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that is much greater than , and thus having in the numerator will make the series diverge by the limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

Determine whether

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges or diverges, and explain why.

Tap to see back →

We can use the alternating series test to show that

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges.

We must have $\small \sin $\frac{1}{n}$\geq 0$ for $\small n\geq 1$ in order to use this test. This is easy to see because $\small $\frac{1}{n}$$ is in $\small \small \left[0,$\frac{\pi}{2}\right]$ for all $\small n\geq 1$ (the values of this sequence are $\small 1,1/2,1/3,1/4,...$ ), and sine is always nonzero whenever sine's argument is in $\small \small \left[0,$\frac{\pi}{2}\right]$ .

Now we must show that

1. $\small \small $\lim_{n\to\infty}$ \sin $\frac{1}{n}$=0$

2. $\small \sin $\frac{1}{n}$$ is a decreasing sequence.

The limit

$\small $\lim_{n\to\infty}$$\frac{1}{n}$=0$

implies that

$\small \small \small $\lim_{n\to\infty}$ \sin $\frac{1}{n}$=\sin 0=0$

so the first condition is satisfied.

We can show that $\small \small \small \sin $\frac{1}{n}$$ is decreasing by taking its derivative and showing that it is less than $\small 0$ for $\small n\geq 1$ :

$\small \small \small \small \small $\frac{d}{dn}$\sin $$\frac{1}{n}$=-$\frac{1}{n^2$}$\cos$\frac{1}{n}$$

The derivative is less than $\small 0$ , because $\small $-$\frac{1}{n^2$}$$ is always less than $\small 0$ , and that $\small \cos $\frac{1}{n}$$ is positive for $\small n\geq 1$ , using a similar argument we used to prove that $\small \small \sin $\frac{1}{n}$\geq 0$ for $\small n\geq 1$ . Since the derivative is less than $\small 0$ , $\small \small \small \sin $\frac{1}{n}$$ is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges, by the alternating series test.

Determine whether the following series converges or diverges:

$\sum_${n=0}^{\infty }$$(-1)^n$($\frac{5}{n}$)$

Tap to see back →

Given just the harmonic series, we would state that the series diverges. However, we are given the alternating harmonic series. To determine whether this series will converge or diverge, we must use the Alternating Series test.

The test states that for a given series $\sum_${n=0}^{\infty $}$a_{n}$$ where $a_${n}=(-1)^$n$(b_{n}$)$ or $a_${n}=(-1)^${n+1}$(b_{n}$)$ where for all n, if and is a decreasing sequence, then $\sum_${n=0}^{\infty $}$a_{n}$$ is convergent.

First, we must evaluate the limit of as n approaches infinity:

The limit equals zero because the numerator of the fraction equals zero as n approaches infinity.

Next, we must determine if is a decreasing sequence. $\frac{1}{n}$< $\frac{1}{n+1}$ , thus the sequence is decreasing.

Because both parts of the test passed, the series is (absolutely) convergent.

Determine whether the series converges or diverges:

$\sum_${n=0}^{\infty }$$(-1)^{n+1}$$\left($\frac{n^4$$}{2n^4$+1}$\right)$

Tap to see back →

To determine whether the series converges or diverges, we must use the Alternating Series test, which states that for

$\sum $a_{n}$$ - and $a_${n}=(-1)^${n+1}$b_{n}$$ where for all n - to converge,

must equal zero and must be a decreasing series.

For our series,

because it behaves like

.

The test fails because $$\frac{1}{2}$\neq0$ so we do not need to check the second condition of the test.

The series is divergent.

For the series: $\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n^n$$}{8^n$}$)$ , determine if the series converge or diverge. If it diverges, choose the best reason.

Tap to see back →

The series given is an alternating series.

Write the three rules that are used to satisfy convergence in an alternating series test.

For $\sum_${n=1}^{\infty}$ $(-1)^{n+1}$ x_n= x_1-x_2+x_3-x_4+...$ :

$\textup{1. All the } x_n \textup{ terms are positive.}$

$\textup{2. The terms must be decreasing: } x_n \geq $x_{n+1}$$

$\textup{3. The limit $}\lim_{n \to \infty}$x_n=0.$

$\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n^n$$}{8^n$}$)=\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n}{8}$)^n$$

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since and instead approaches infinity.

The correct answer is:

$\textup{Diverge, $since}\lim_{n \to \infty}$x_n\neq 0.$

Which of the following series does not converge?

Tap to see back →

We can show that the series $\sum_${i=0}^{\infty}$ $$\frac{n!}{n^2$ cos(n)}$$ diverges using the ratio test.

$L = $\lim_{n ->\infty }$ $$\frac{a_{n+1}$$$}{a_{n}$} = $\lim_{n ->\infty }$ $\left[\left($\frac{(n+1)!}{(n+1)^2$ cos (n+1)}$ \div $$\frac{n!}{n^2$ cos (n) }$\right)\right]$

$= $\lim_{n ->\infty}$ $$\frac{n^2$$(n+1)cos(n)}{(n+1)^2cos$(n+1)}$ = $\lim_{n ->\infty}$ $$\frac{n^2cos$(n)}{(n+1)cos(n+1)}$$

will dominate over since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that is much greater than , and thus having in the numerator will make the series diverge by the limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

Determine whether

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges or diverges, and explain why.

Tap to see back →

We can use the alternating series test to show that

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges.

We must have $\small \sin $\frac{1}{n}$\geq 0$ for $\small n\geq 1$ in order to use this test. This is easy to see because $\small $\frac{1}{n}$$ is in $\small \small \left[0,$\frac{\pi}{2}\right]$ for all $\small n\geq 1$ (the values of this sequence are $\small 1,1/2,1/3,1/4,...$ ), and sine is always nonzero whenever sine's argument is in $\small \small \left[0,$\frac{\pi}{2}\right]$ .

Now we must show that

1. $\small \small $\lim_{n\to\infty}$ \sin $\frac{1}{n}$=0$

2. $\small \sin $\frac{1}{n}$$ is a decreasing sequence.

The limit

$\small $\lim_{n\to\infty}$$\frac{1}{n}$=0$

implies that

$\small \small \small $\lim_{n\to\infty}$ \sin $\frac{1}{n}$=\sin 0=0$

so the first condition is satisfied.

We can show that $\small \small \small \sin $\frac{1}{n}$$ is decreasing by taking its derivative and showing that it is less than $\small 0$ for $\small n\geq 1$ :

$\small \small \small \small \small $\frac{d}{dn}$\sin $$\frac{1}{n}$=-$\frac{1}{n^2$}$\cos$\frac{1}{n}$$

The derivative is less than $\small 0$ , because $\small $-$\frac{1}{n^2$}$$ is always less than $\small 0$ , and that $\small \cos $\frac{1}{n}$$ is positive for $\small n\geq 1$ , using a similar argument we used to prove that $\small \small \sin $\frac{1}{n}$\geq 0$ for $\small n\geq 1$ . Since the derivative is less than $\small 0$ , $\small \small \small \sin $\frac{1}{n}$$ is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges, by the alternating series test.

Determine whether the following series converges or diverges:

$\sum_${n=0}^{\infty }$$(-1)^n$($\frac{5}{n}$)$

Tap to see back →

Given just the harmonic series, we would state that the series diverges. However, we are given the alternating harmonic series. To determine whether this series will converge or diverge, we must use the Alternating Series test.

The test states that for a given series $\sum_${n=0}^{\infty $}$a_{n}$$ where $a_${n}=(-1)^$n$(b_{n}$)$ or $a_${n}=(-1)^${n+1}$(b_{n}$)$ where for all n, if and is a decreasing sequence, then $\sum_${n=0}^{\infty $}$a_{n}$$ is convergent.

First, we must evaluate the limit of as n approaches infinity:

The limit equals zero because the numerator of the fraction equals zero as n approaches infinity.

Next, we must determine if is a decreasing sequence. $\frac{1}{n}$< $\frac{1}{n+1}$ , thus the sequence is decreasing.

Because both parts of the test passed, the series is (absolutely) convergent.

Determine whether the series converges or diverges:

$\sum_${n=0}^{\infty }$$(-1)^{n+1}$$\left($\frac{n^4$$}{2n^4$+1}$\right)$

Tap to see back →

To determine whether the series converges or diverges, we must use the Alternating Series test, which states that for

$\sum $a_{n}$$ - and $a_${n}=(-1)^${n+1}$b_{n}$$ where for all n - to converge,

must equal zero and must be a decreasing series.

For our series,

because it behaves like

.

The test fails because $$\frac{1}{2}$\neq0$ so we do not need to check the second condition of the test.

The series is divergent.

For the series: $\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n^n$$}{8^n$}$)$ , determine if the series converge or diverge. If it diverges, choose the best reason.

Tap to see back →

The series given is an alternating series.

Write the three rules that are used to satisfy convergence in an alternating series test.

For $\sum_${n=1}^{\infty}$ $(-1)^{n+1}$ x_n= x_1-x_2+x_3-x_4+...$ :

$\textup{1. All the } x_n \textup{ terms are positive.}$

$\textup{2. The terms must be decreasing: } x_n \geq $x_{n+1}$$

$\textup{3. The limit $}\lim_{n \to \infty}$x_n=0.$

$\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n^n$$}{8^n$}$)=\sum_${n=0}^{\infty}$ $(-1)^n$ $($\frac{n}{8}$)^n$$

The first and second conditions are satisfied since the terms are positive and are decreasing after each term.

However, the third condition is not valid since and instead approaches infinity.

The correct answer is:

$\textup{Diverge, $since}\lim_{n \to \infty}$x_n\neq 0.$

Which of the following series does not converge?

Tap to see back →

We can show that the series $\sum_${i=0}^{\infty}$ $$\frac{n!}{n^2$ cos(n)}$$ diverges using the ratio test.

$L = $\lim_{n ->\infty }$ $$\frac{a_{n+1}$$$}{a_{n}$} = $\lim_{n ->\infty }$ $\left[\left($\frac{(n+1)!}{(n+1)^2$ cos (n+1)}$ \div $$\frac{n!}{n^2$ cos (n) }$\right)\right]$

$= $\lim_{n ->\infty}$ $$\frac{n^2$$(n+1)cos(n)}{(n+1)^2cos$(n+1)}$ = $\lim_{n ->\infty}$ $$\frac{n^2cos$(n)}{(n+1)cos(n+1)}$$

will dominate over since it's a higher order term. Clearly, L will not be less than, which is necessary for absolute convergence.

Alternatively, it's clear that is much greater than , and thus having in the numerator will make the series diverge by the limit test (since the terms clearly don't converge to zero).

The other series will converge by alternating series test, ratio test, geometric series, and comparison tests.

Determine whether

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges or diverges, and explain why.

Tap to see back →

We can use the alternating series test to show that

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges.

We must have $\small \sin $\frac{1}{n}$\geq 0$ for $\small n\geq 1$ in order to use this test. This is easy to see because $\small $\frac{1}{n}$$ is in $\small \small \left[0,$\frac{\pi}{2}\right]$ for all $\small n\geq 1$ (the values of this sequence are $\small 1,1/2,1/3,1/4,...$ ), and sine is always nonzero whenever sine's argument is in $\small \small \left[0,$\frac{\pi}{2}\right]$ .

Now we must show that

1. $\small \small $\lim_{n\to\infty}$ \sin $\frac{1}{n}$=0$

2. $\small \sin $\frac{1}{n}$$ is a decreasing sequence.

The limit

$\small $\lim_{n\to\infty}$$\frac{1}{n}$=0$

implies that

$\small \small \small $\lim_{n\to\infty}$ \sin $\frac{1}{n}$=\sin 0=0$

so the first condition is satisfied.

We can show that $\small \small \small \sin $\frac{1}{n}$$ is decreasing by taking its derivative and showing that it is less than $\small 0$ for $\small n\geq 1$ :

$\small \small \small \small \small $\frac{d}{dn}$\sin $$\frac{1}{n}$=-$\frac{1}{n^2$}$\cos$\frac{1}{n}$$

The derivative is less than $\small 0$ , because $\small $-$\frac{1}{n^2$}$$ is always less than $\small 0$ , and that $\small \cos $\frac{1}{n}$$ is positive for $\small n\geq 1$ , using a similar argument we used to prove that $\small \small \sin $\frac{1}{n}$\geq 0$ for $\small n\geq 1$ . Since the derivative is less than $\small 0$ , $\small \small \small \sin $\frac{1}{n}$$ is a decreasing sequence. Now we have shown that the two conditions are satisfied, so we have proven that

$\small \sum_${n=1}^{\infty}$ $(-1)^n$ \sin$\frac{1}{n}$$

converges, by the alternating series test.

Determine whether the following series converges or diverges:

$\sum_${n=0}^{\infty }$$(-1)^n$($\frac{5}{n}$)$

Tap to see back →

Given just the harmonic series, we would state that the series diverges. However, we are given the alternating harmonic series. To determine whether this series will converge or diverge, we must use the Alternating Series test.

The test states that for a given series $\sum_${n=0}^{\infty $}$a_{n}$$ where $a_${n}=(-1)^$n$(b_{n}$)$ or $a_${n}=(-1)^${n+1}$(b_{n}$)$ where for all n, if and is a decreasing sequence, then $\sum_${n=0}^{\infty $}$a_{n}$$ is convergent.

First, we must evaluate the limit of as n approaches infinity:

The limit equals zero because the numerator of the fraction equals zero as n approaches infinity.

Next, we must determine if is a decreasing sequence. $\frac{1}{n}$< $\frac{1}{n+1}$ , thus the sequence is decreasing.

Because both parts of the test passed, the series is (absolutely) convergent.