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AP Calculus BC : Series of Constants

Study concepts, example questions & explanations for AP Calculus BC

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2 Diagnostic Tests 86 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Alternating Series With Error Bound

Determine whether the series converges or diverges:

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

Possible Answers:

The series is divergent.

The series may be convergent, divergent, or conditionally convergent.

The series is conditionally convergent.

The series is (absolutely) convergent.

Correct answer:

The series is divergent.

Explanation:

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 $b_{n} \geq 0$ for all n - to converge,

$\lim_{n\rightarrow \infty }b_{n}$ must equal zero and $b_{n}$ must be a decreasing series.

For our series,

$\lim_{n\rightarrow \infty }b_{n}=\lim_{n\rightarrow \infty }\frac{n^4}{2n^4+1}=\frac{1}{2}$

because it behaves like

$\lim_{n\rightarrow \infty }\left(\frac{1}{2}\right)\frac{n^4}{n^4}=\frac{1}{2}$ .

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.

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Example Question #71 : Ap Calculus Bc

Which of the following series does not converge?

Possible Answers:

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

$\sum_{i=0}^{\infty} \frac{n^2 ln (n)}{n!}$

$\sum_{i=0}^{\infty} 0.9999999999999^n$

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

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

Correct answer:

$\sum_{i=0}^{\infty} \frac{n!}{n^2 cos(n)}$

Explanation:

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

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

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

n^2 will dominate over (n+1) 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 n^2 , and thus having in the numerator will make the series diverge by the $n^{th}$ 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.

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Example Question #1 : Concepts Of Convergence And Divergence

One of the following infinite series CONVERGES. Which is it?

Possible Answers:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$

None of the others converge.

$\sum_{k=1}^{\infty} (-1)^k sin(k)$

$\sum_{k=2}^{\infty} \frac{x}{ln(x)}$

$\sum_{k=1}^{\infty} \frac{1}{k}$

Correct answer:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$

Explanation:

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ converges due to the comparison test.

We start with the equation $\frac{1}{k^2} =\frac{1}{k^2}$ . Since $sin(k) \leq 1$ for all values of k, we can multiply both side of the equation by the inequality and get $\frac{sin(k)}{k^2} \leq \frac{1}{k^2}$ for all values of k. Since $\sum_{k=1}^{\infty} \frac{1}{k^2}$ is a convergent p-series with p=2 , $\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ hence also converges by the comparison test.

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Example Question #72 : Ap Calculus Bc

Determine the nature of convergence of the series having the general term:

$\frac{8}{\sqrt{n(n+1)(n+2)}}$

Possible Answers:

$\frac{1}{2}$

The series is convergent.

The series is divergent.

$\frac{\pi}{7}$

$\frac{\pi}{5}$

Correct answer:

The series is convergent.

Explanation:

We will use the Limit Comparison Test to establish this result.

We need to note that the following limit

$\frac{\frac{8}{\sqrt{n(n+1)(n+2)}}}{\frac{8}{n^{3/2}}}}$

goes to 1 as n goes to infinity.

Therefore the series have the same nature. They either converge or diverge at the same time.

We will focus on the series:

$\frac{1}{n^{3/2}}$ .

We know that this series is convergent because it is a p-series. (Remember that

$\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges if p>1 and we have p=3/2 which is greater that one in this case)

By the Limit Comparison Test, we deduce that the series is convergent, and that is what we needed to show.

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Example Question #11 : Series Of Constants

Determine if the series converges or diverges. You do not need to find the sum.

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

Possible Answers:

Converges

There is not enough information to decide convergence.

Neither converges nor diverges.

Diverges

Conditionally converges.

Correct answer:

Converges

Explanation:

We can compare this to the series $\sum_{n = 1}^{\infty } \frac{1}{n^{3}}$ which we know converges by the p-series test.

To figure this out, let's first compare $n^{3} + 2$ to $n^{3}$ . For any number n, $n^{3} + 2$ will be larger than $n^{3}$ .

There is a rule in math that if you take the reciprocal of each term in an inequality, you are allowed to flip the signs.

Thus, $n^{3} + 2 > n^{3}}$ turns into

$\frac{1}{n^{3}} > \frac{1}{n^{3} + 2}$ .

And so, because $\frac{1}{n ^{3}}$ converges, thus our series also converges.

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Example Question #73 : Ap Calculus Bc

Which of the following tests will help determine whether $\sum_{x=1}^{\infty}\frac{1}{x}$ is convergent or divergent, and why?

Possible Answers:

Integral Test: The improper integral determines that the harmonic series diverge.

Root Test: Since the limit as approaches to infinity is zero, the series is convergent.

P-Series Test: The summation converges since p=1 .

$\textup{}$ Divergence Test: Since limit of the series approaches zero, the series must converge.

$\textup{}$ Nth Term Test: The series diverge because the limit as goes to infinity is zero.

Correct answer:

Integral Test: The improper integral determines that the harmonic series diverge.

Explanation:

The series $\sum_{x=1}^{\infty}\frac{1}{x}$ is a harmonic series.

The Nth term test and the Divergent test may not be used to determine whether this series converges, since this is a special case. The root test also does not apply in this scenario.

According the the P-series Test, $\sum_{x=1}^{\infty}\frac{1}{x^p}$ must converge only if p>1 . Therefore this could be a valid test, but a wrong definition as the answer choice since the series diverge for $p \leq1$ .

This leaves us with the Integral Test.

$\int_{1}^{\infty}\frac{1}{x}= \lim_{b \to \infty}\int_{1}^{b}\frac{1}{x}dx=\lim_{b \to \infty}( ln(b)-ln(1))=\infty$

Since the improper integral diverges, so does the series.

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Example Question #72 : Polynomial Approximations And Series

Does the series $\sum_{n=0}^{\infty }\frac{(-1)^n}{n}$ converge conditionally, absolutely, or diverge?

Possible Answers:

Does not exist.

Converge Conditionally.

Cannot tell with the given information.

Converge Absolutely.

Diverges.

Correct answer:

Converge Conditionally.

Explanation:

The series converges conditionally.

The absolute values of the series $\sum_{n=0}^{\infty }\left |\frac{(1)^n}{n} \right |$ is a divergent p-series with $p\leq 1$ .

However, the the limit of the sequence $\lim_{n\rightarrow \infty }\frac{(-1)^n}{n}=0$ and it is a decreasing sequence.

Therefore, by the alternating series test, the series converges conditionally.

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Example Question #1 : P Series

True or False, a -series cannot be tested conclusively using the ratio test.

Possible Answers:

False

True

Correct answer:

True

Explanation:

We cannot test for convergence of a -series using the ratio test. Observe,

For the series $\sum_{n=1}^\infty \frac{1}{n^p}$ ,

$\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}| = \lim_{n \to \infty} | \frac{n^p}{(n+1)^p}| = \lim_{n \to \infty} (\frac{n}{n+1})^p = 1^p =1$ .

Since this limit is regardless of the value for , the ratio test is inconclusive.

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Example Question #2881 : Calculus Ii

Determine if the following series is divergent, convergent or neither.

$\sum_{n=1}^{\infty} \frac{(-9)^n}{5^{3n+2}(n+1)}$

Possible Answers:

Divergent

Convergent

Neither

Both

Inconclusive

Correct answer:

Convergent

Explanation:

In order to figure out if

$\sum_{n=1}^{\infty} \frac{(-9)^n}{5^{3n+2}(n+1)}$

is divergent, convergent or neither, we need to use the ratio test.

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n} \right |$

Then if

L<1 , the series is absolutely convergent.

L>1 , the series is divergent.

L=1 , the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply the ratio test to our problem.

Let

$a_n=\frac{(-9)^n}{5^{3n+2}(n+1)}$

and

$a_{n+1}=\frac{(-9)^{n+1}}{5^{3(n+1)+2}((n+1)+1)}$

$=\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}}{\frac{(-9)^n}{5^{3n+2}(n+1)}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{(-9)^{n+1}5^{3n+2}(n+1)}{(-9)^n5^{3n+5}(n+2)} \right |$

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{(-9)^{1}(n+1)}{5^{3}(n+2)} \right |$

$=\frac{9}{125}\lim_{n\rightarrow \infty}\left | \frac{n+1}{n+2} \right |$

$=\frac{9}{125}*1=\frac{9}{125}$ .

Since

$\frac{9}{125}<1$ .

We have sufficient evidence to conclude that the series is convergent.

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Example Question #81 : Polynomial Approximations And Series

Determine if the following series is divergent, convergent or neither.

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

Possible Answers:

Neither

Inconclusive

Convergent

Divergent

Both

Correct answer:

Divergent

Explanation:

In order to figure if

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

is convergent, divergent or neither, we need to use the ratio test.

Remember that the ratio test is as follows.

Suppose we have a series $\sum a_n$ . We define,

$L=\lim_{n\rightarrow \infty}\left | \frac{a_{n+1}}{a_n} \right |$

Then if

L<1 , the series is absolutely convergent.

L>1 , the series is divergent.

L=1 , the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply the ratio test to our problem.

Let

$a_n=\frac{n!}{2^{n+1}}$

and

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

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)!}{2^{n+2}}}{\frac{n!}{2^{n+1}}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{2^{n+1}(n+1)!}{2^{n+2}n!} \right |$ .

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{n+1}{2} \right |$

$=\frac{1}{2}\lim_{n\rightarrow \infty}\left | n+1 \right |$

$=\frac{1}{2}*\infty=\infty$ .

Since $\infty>1$ ,

we have sufficient evidence to conclude that the series is divergent.

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AP Calculus BC : Series of Constants

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Alternating Series With Error Bound

Example Question #71 : Ap Calculus Bc

Example Question #1 : Concepts Of Convergence And Divergence

Example Question #72 : Ap Calculus Bc

Example Question #11 : Series Of Constants

Example Question #73 : Ap Calculus Bc

Example Question #72 : Polynomial Approximations And Series

Example Question #1 : P Series

Example Question #2881 : Calculus Ii

Example Question #81 : Polynomial Approximations And Series

All AP Calculus BC Resources

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