Calculus 2 : Alternating Series

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #3001 : Calculus Ii

By definition, an Alternating Series is a series of the form-

Possible Answers:

\displaystyle \sum_{k=k_0}^{\infty} (-1)^k \cdot a_k

\displaystyle \sum_{k=k_0}^{\infty} [(-1)^k\cdot a_{k+1} + (-1)^{k+1} \cdot a_k]

\displaystyle \sum_{k=k_0}^{\infty} (-1)^k + a_k \cdot a_{k+1}

None of the other answers

Correct answer:

\displaystyle \sum_{k=k_0}^{\infty} (-1)^k \cdot a_k

Explanation:

This type of series we can frequently check for convergence/divergence using the Alternating Series Test.

\displaystyle \sum_{n=1}^\infty(-1)^{n+1}a_n=a_1-a_2+a_3-a_4+...

The terms with an odd value for \displaystyle n become negative since \displaystyle (-1)^n=-1 and the terms with an even value for \displaystyle n are positive. This creates the alternating signs to occur within the sum.

Example Question #3002 : Calculus Ii

Differentiate the following function.

\displaystyle f(x)=x^4+\frac{x}{3x^2}+9

Possible Answers:

\displaystyle f'(x)=\frac{1}{4}x^2+3x

\displaystyle f'(x)=x^3+999

\displaystyle f'(x)=4x^3-\frac{1}{3x^2}

\displaystyle f'(x)=4x^3+\frac{1}{3x^3}

Correct answer:

\displaystyle f'(x)=4x^3-\frac{1}{3x^2}

Explanation:

To differentiate the function we will need to use the Power Rule which states:

\displaystyle f(x)=ax^n \rightarrow f'(x)=nax^{n-1}

Looking at our function we can first simplify the equation.

\displaystyle f(x)=x^4+\frac{x}{3x^2}+9=x^4+\frac{1}{3x}+9

Applying the Power Rule we get:

\displaystyle f'(x)=4x^3-\frac{1}{3x^2}

Example Question #4 : P Series

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

Possible Answers:

Diverges.

Cannot tell with the given information.

Does not exist.

Converge Absolutely.

Converge Conditionally.

Correct answer:

Converge Conditionally.

Explanation:

The series converges conditionally.

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

However, the the limit of the sequence \displaystyle \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.    

Example Question #61 : Polynomial Approximations And Series

Find the interval of convergence of \displaystyle x for the series \displaystyle \sum_{n=1}^{\infty }\frac{3^nx^n}{n!}.

Possible Answers:

\displaystyle \left ( \frac{-1}{3}, \frac{1}{3} \right )

\displaystyle \left ( -3, 3\right )

\displaystyle (-1,0)

\displaystyle \left [ \frac{-1}{3}, \frac{1}{3} \right ]

\displaystyle \left ( -\infty , \infty \right )

Correct answer:

\displaystyle \left ( -\infty , \infty \right )

Explanation:

Using the root test, 

\displaystyle \lim_{n\rightarrow \infty }\left | \frac{a_{n+1}}{a_n} \right |=\lim_{n\rightarrow \infty }\left |\frac{3^{n+1} x^{n+1}}{(n+1)!} \cdot \frac{n!}{3^nx^n} \right |=\left | 3x\right |\lim_{n\rightarrow \infty }\left | \frac{1}{n+1} \right |=0

Because 0 is always less than 1, the root test shows that the series converges for any value of x. 

Therefore, the interval of convergence is:

\displaystyle (-\infty, \infty)

Example Question #1 : Alternating Series With Error Bound

Determine whether 

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

converges or diverges, and explain why.

Possible Answers:

Divergent, by the comparison test.

More tests are needed.

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

Divergent, by the test for divergence.

Convergent, by the alternating series test. 

Correct answer:

Convergent, by the alternating series test. 

Explanation:

We can use the alternating series test to show that

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

converges.

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

Now we must show that

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

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

The limit 

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

implies that 

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

so the first condition is satisfied.

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

\displaystyle \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 \displaystyle \small 0, because \displaystyle \small -\frac{1}{n^2} is always less than \displaystyle \small 0, and that \displaystyle \small \cos \frac{1}{n} is positive for \displaystyle \small n\geq 1, using a similar argument we used to prove that \displaystyle \small \small \sin \frac{1}{n}\geq 0 for \displaystyle \small n\geq 1. Since the derivative is less than \displaystyle \small 0\displaystyle \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 

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

converges, by the alternating series test.

Example Question #71 : Polynomial Approximations And Series

For the series:  \displaystyle \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.

Possible Answers:

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

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

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

\displaystyle \textup{Converge, since}\lim_{n \to \infty}x_n\neq 0.

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

Correct answer:

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

Explanation:

The series given is an alternating series.  

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

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

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

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

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

\displaystyle \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 \displaystyle \lim_{n \to \infty}x_n\neq 0 and instead approaches infinity.

The correct answer is:

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

Example Question #1 : Alternating Series

Write a series expression for \displaystyle n terms of the following sequence.

\displaystyle 1-2+4-8+16-32+64-128...

Possible Answers:

\displaystyle \sum_{k=0}^{n} 2^k

\displaystyle \sum_{k=0}^{n} k^2

This sequence can't be represented as a series.

\displaystyle \sum_{k=0}^{n} 2k

\displaystyle \sum_{k=0}^{n} (-2)^k

Correct answer:

\displaystyle \sum_{k=0}^{n} (-2)^k

Explanation:

If we look at this sequence

\displaystyle 1-2+4-8+16-32+64-128...

The first thing we should notice is that it is alternating from positive to negative. This means that we will have

\displaystyle (-1)^k.

The second thing we should notice is that the sequence is increasing in powers of 2.

Thus we will also have

\displaystyle 2^k.

Now we can combine these statements and write them in terms of a series.

\displaystyle \sum_{k=0}^{n} (-1)^k*2^k

We can now simplify this into 

\displaystyle \sum_{k=0}^{n} (-2)^k.

Example Question #8 : Alternating Series

Determine whether the series is convergent or divergent:

\displaystyle \sum_{n=0}^{\infty }(-1)^n\left(\frac{n^3}{3n^3+n^2+1}\right)

Possible Answers:

The series is divergent.

The series is (absolutely) convergent.

The series is conditionally convergent.

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

Correct answer:

The series is divergent.

Explanation:

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

\displaystyle \sum a_{n} and \displaystyle a_{n}=(-1)^n(b_{n}),

where \displaystyle b_{n}\geq n for all n, if \displaystyle \lim_{n\rightarrow \infty }b_{n}=0 and \displaystyle {b_{n}} is a decreasing sequence, then the series is convergent.

First, we must identify \displaystyle {b_{n}}, which is \displaystyle \frac{n^3}{3n^3+n^2+1}. When we take the limit of \displaystyle {b_{n}} as n approaches infinity, we get

\displaystyle \lim_{n \to \infty }\frac{n^3}{3n^3+n^2+1}=\lim_{n \to \infty }\left(\frac{n^3}{n^3}\right)\left(\frac{1}{3+n^{-1}+n^{-3}}\right)=\frac{1}{3}\neq0

Notice that for the limit, the negative power terms go to zero, so we are left with something that does not equal zero.

Thus, the series is divergent because the test fails. 

Example Question #1 : Alternating Series With Error Bound

Determine whether the series converges or diverges:

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

Possible Answers:

The series is divergent.

The series is conditionally convergent.

The series may be convergent, divergent, or 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 

\displaystyle \sum a_{n} - and \displaystyle a_{n}=(-1)^{n+1}b_{n} where \displaystyle b_{n} \geq 0 for all n - to converge, 

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

For our series, 

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

because it behaves like 

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

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

The series is divergent.

 

 

Example Question #1 : Alternating Series

Determine if the following series is convergent or divergent:

\displaystyle \sum_{n=1}^{\infty} (-1)^{n}\frac{1}{n+n^2}

Possible Answers:

Divergent according to the ratio test

Divergent according to the alternating series test

Inconclusive according to the alternating series test

Convergent according to the alternating series test

Correct answer:

Convergent according to the alternating series test

Explanation:

This is an alternating series.

An alternating series can be identified because terms in the series will “alternate” between + and –, because of \displaystyle (-1)^{n}

Note: Alternating Series Test can only show convergence. It cannot show divergence.

If the following 2 tests are true, the alternating series converges.

\displaystyle a_{n} = (-1)^{n} * b_{n}

  1. \displaystyle \lim_{n \rightarrow \infty} b_{n} = 0
  2. {\displaystyle b_{n}} is a decreasing sequence, or in other words \displaystyle b_{n=1}' < 0

Solution:

 

\displaystyle b_{n}=\frac{1}{n+n^2}

1.

\displaystyle \lim_{n \rightarrow \infty}b_{n}

\displaystyle \lim_{n \rightarrow \infty}\frac{1}{n+n^2}=0

2.

\displaystyle b_{n}'=(\frac{1}{n+n^2})'

\displaystyle b_{n}'=\frac{-(n^{2}+n)'}{(n^2+n)^2}

\displaystyle b_{n}'=\frac{-(2n+1)}{(n^2+n)^{2}}

\displaystyle b_{1}'=\frac{-(2+1)}{(1+1)^2}=\frac{-3}{4} < 0

Since the 2 tests pass, this series is convergent.

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