AP Calculus BC - Series of Constants

Assuming that $a_{n}=n^{\alpha}(n+1)$ , $\alpha \ge 2$ . Using the ratio test, what can we say about the series:

$\sum_{n=1}^{\infty} \frac{1}{a_{n}}$

We cannot conclude when we use the ratio test.

It is convergent.

$\infty$

$-\infty$

Explanation

As required by this question we will have to use the ratio test. $L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

To do so, we will need to compute : $\frac{a_{n+1}}{a_{n}}$ . In our case:

$a_{n}=\frac{1}{a_{n}}$

Therefore

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}$ .

We know that $\lim_{n\rightarrow \infty }\frac{n}{n+1}=1$

This means that

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}=1\forall\quad \alpha$

Since L=1 by the ratio test, we can't conclude about the convergence of the series.

Assuming that $a_{n}=n^{\alpha}(n+1)$ , $\alpha \ge 2$ . Using the ratio test, what can we say about the series:

$\sum_{n=1}^{\infty} \frac{1}{a_{n}}$

We cannot conclude when we use the ratio test.

It is convergent.

$\infty$

$-\infty$

Explanation

To do so, we will need to compute : $\frac{a_{n+1}}{a_{n}}$ . In our case:

$a_{n}=\frac{1}{a_{n}}$

Therefore

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}$ .

We know that $\lim_{n\rightarrow \infty }\frac{n}{n+1}=1$

This means that

$\frac{a_{n+1}}{a_{n}}=(\frac{n}{n+1})^{\alpha}\frac{n+1}{n+2}=1\forall\quad \alpha$

Since L=1 by the ratio test, we can't conclude about the convergence of the series.

Determine whether

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

converges or diverges, and explain why.

Convergent, by the alternating series test.

Convergent, by the $\small p$ -series test.

Divergent, by the test for divergence.

Divergent, by the comparison test.

More tests are needed.

Explanation

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.

We consider the following series:

$\sum_{n=0}^{\infty} \frac{n^3+1}{n^4}$$

Determine the nature of the convergence of the series.

The series is divergent.

$\frac{1}{2}$

$\frac{3}4{}$

Explanation

We will use the comparison test to prove this result. We must note the following:

$\frac{n^3+1}{n^4}$ is positive.

We have all natural numbers n:

$n^4\ge n^3$ , this implies that

$n^4\ge n^3 \ge n$ .

Inverting we get :

$\frac{1}{n} <\frac{n^3+1}{n^4}$

Summing from 1 to $\infty$ , we have

$\sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n^3+1}{n^4}$

We know that the $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. Therefore by the comparison test:

$\sum_{n=1}^{\infty}\frac{n^3}{n^4+1}$

is divergent

We consider the following series:

$\sum_{n=0}^{\infty} \frac{n^3+1}{n^4}$$

Determine the nature of the convergence of the series.

The series is divergent.

$\frac{1}{2}$

$\frac{3}4{}$

Explanation

We will use the comparison test to prove this result. We must note the following:

$\frac{n^3+1}{n^4}$ is positive.

We have all natural numbers n:

$n^4\ge n^3$ , this implies that

$n^4\ge n^3 \ge n$ .

Inverting we get :

$\frac{1}{n} <\frac{n^3+1}{n^4}$

Summing from 1 to $\infty$ , we have

$\sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n^3+1}{n^4}$

We know that the $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. Therefore by the comparison test:

$\sum_{n=1}^{\infty}\frac{n^3}{n^4+1}$

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.

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

$\textup{Converge, since}\lim_{n \to \infty}x_n eq 0.$

$\textup{Converge, since all Alternating Series Test rules are satisfied.}$

$\textup{Diverges, since the terms are not progressively decreasing}.$

$\textup{Diverges, by the Geometric Series Test.}$

Explanation

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 $\lim_{n \to \infty}x_n eq 0$ and instead approaches infinity.

The correct answer is:

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

We consider the following series:

$\sum_{n=0}^{\infty} \frac{n}{n^2+1}$$

Determine the nature of the convergence of the series.

The series is divergent.

$\frac{1}{2}$

$\frac{3}4{}$

Explanation

We will use the Comparison Test to prove this result. We must note the following:

$\frac{n}{n^2+1}$ is positive.

We have all natural numbers n:

$n^2\ge n$ , this implies that

$n^2+1\ge n+1 > n$ .

Inverting we get :

$\frac{1}{n} <\frac{n}{n^2+1}$

Summing from 1 to $\infty$ , we have

$\sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n}{n^2+1}$

We know that the $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. Therefore by the Comparison Test:

$\sum_{n=1}^{\infty}\frac{n}{n^2+1}$ 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.

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

$\textup{Converge, since}\lim_{n \to \infty}x_n eq 0.$

$\textup{Converge, since all Alternating Series Test rules are satisfied.}$

$\textup{Diverges, since the terms are not progressively decreasing}.$

$\textup{Diverges, by the Geometric Series Test.}$

Explanation

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 $\lim_{n \to \infty}x_n eq 0$ and instead approaches infinity.

The correct answer is:

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

Determine whether

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

converges or diverges, and explain why.

Convergent, by the alternating series test.

Convergent, by the $\small p$ -series test.

Divergent, by the test for divergence.

Divergent, by the comparison test.

More tests are needed.

Explanation

We can use the alternating series test to show that

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

converges.

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 $\small n\geq 1$ :

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

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

converges, by the alternating series test.

We consider the following series:

$\sum_{n=0}^{\infty} \frac{n}{n^2+1}$$

Determine the nature of the convergence of the series.

The series is divergent.

$\frac{1}{2}$

$\frac{3}4{}$

Explanation

We will use the Comparison Test to prove this result. We must note the following:

$\frac{n}{n^2+1}$ is positive.

We have all natural numbers n:

$n^2\ge n$ , this implies that

$n^2+1\ge n+1 > n$ .

Inverting we get :

$\frac{1}{n} <\frac{n}{n^2+1}$

Summing from 1 to $\infty$ , we have

$\sum_{n=1}^{\infty}\frac{1}{n} < \sum_{n=1}^{\infty}\frac{n}{n^2+1}$

We know that the $\sum_{n=1}^{\infty}\frac{1}{n}$ is divergent. Therefore by the Comparison Test:

$\sum_{n=1}^{\infty}\frac{n}{n^2+1}$ is divergent.

Series of Constants

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