Polynomial Approximations and Series - AP Calculus BC

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Question

Which of the following series does not converge?

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Answer

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 $n^{th}$ 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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Question

Determine whether

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

converges or diverges, and explain why.

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Answer

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.

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Question

Determine whether the following series converges or diverges:

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

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Answer

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 $b_{n}\geq0$ for all n, if $\lim_{n\rightarrow \infty }b_{n}=0$ and $b_{n}$ is a decreasing sequence, then $\sum_{n=0}^{\infty }a_{n}$ is convergent.

First, we must evaluate the limit of $b_{n}$ as n approaches infinity:

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

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

Next, we must determine if $b_{n}$ 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.

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Question

Consider: $\sum_{n=0}^{\infty} \left (\frac{\sqrt 2}{2}\right )^n$ . Will the series converge or diverge? If converges, where does this coverge to?

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Answer

This is a geometric series. Use the following formula, where is the first term of the series, and is the ratio that must be less than 1. If is greater than 1, the series diverges.

$\frac{a}{1-r} = \frac{1}{1-\frac{\sqrt2}{2}}= \frac{1}{\frac{2-\sqrt2}{2}} = \frac{2}{2-\sqrt2}$

Rationalize the denominator.

$\frac{2}{2-\sqrt2} \cdot \frac{2+\sqrt2}{2+\sqrt2} = \frac{2(2+\sqrt2)}{4-2}= 2+\sqrt2$

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Question

Consider the following summation: $1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...$ . Does this converge or diverge? If it converges, where does it approach?

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Answer

The problem can be reconverted using a summation symbol, and it can be seen that this is geometric.

$1+\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...=\sum_{n=0}^{\infty} \left ( \frac{1}{4} \right )^n$

Since the ratio is less than 1, this series will converge. The formula for geometric series is:

$\frac{a}{1-r}$

where is the first term, and is the common ratio. Substitute these values and solve.

$\frac{a}{1-r} = \frac{1}{1-\frac{1}{4}} = \frac{1}{\frac{3}{4}} = \frac{4}{3}$

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Question

A worm crawls up a wall during the day and slides down slowly during the night. The first day the worm crawls one meter up the wall. The first night the worm slides down a third of a meter. The second day the worm regains one third of the lost progress and slides down one third of that distance regained on the second night. This pattern of motion continues...

Which of the following is a geometric sum representing the distance the worm has travelled after 12-hour periods of motion? (Assuming day and night are both 12 hour periods).

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Answer

The sum must be alternating, and after one period you should have the worm at 1m. After two periods, the worm should be at 2/3m. There is only one sum for which that is true.

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Question

Determine what the following series converges to, and whether the series is Convergent, Divergent or Neither.

$\sum_{n=0}^{\infty} \frac{n^2\ 2^n}{(n+1)\ 3^n}$

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Answer

To determine if

$\sum_{n=0}^{\infty} \frac{n^2\ 2^n}{(n+1)\ 3^n}$

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

The ratio test is as follows.

Suppose we a series $\sum a_n$ . Then we define,

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

If

the series is absolutely convergent (and hence convergent).

the series is divergent.

the series may be divergent, conditionally convergent, or absolutely convergent.

Now lets apply this to our situtation.

Let

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

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

When we evaluate the limit, we get.

$L=\frac{2}{3}$ .

Since , we have sufficient evidence to conclude that the series is Convergent.

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Question

Determine whether the following series converges or diverges. If it converges, what does it converge to?

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

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Answer

First, we reduce the series into a simpler form.

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

We know this series converges because

$\lim_{n \rightarrow \infty}(\frac{1}{4})^{n}=0$

By the Geometric Series Theorem, the sum of this series is given by

$\sum_{n=0}^{\infty}(\frac{1}{4})^{n}= \frac{(\frac{1}{4})^0}{1-\frac{1}{4}}= \frac{4}{3}$

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Question

Calculate the sum of a geometric series with the following values:, $a_{1}=4$ , $r=\frac{1}{3}$ . Round the answer to the nearest integer.

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Answer

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$S_{15}=\frac{a_{1}(1-r^n)}{1-r}$

$S_{15}=\frac{4(1-(\frac{1}{3})^{15})}{1-\frac{1}{3}}$

$S_{15}=\frac{43(1-\frac{1}{3^{15}})}{2}$

$S_{15}=6*(1-\frac{1}{3^{15}})$

$S_{15}=5.999...$

Rounding, $S_{15}=6$

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Question

Calculate the sum of a geometric series with the following values:

, $a_{1}=11$ , $r=\frac{2}{20}$ ,

rounded to the nearest integer.

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Answer

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

For this question, we are given all of the information we need.

Solution:

$S_{10}=\frac{a_{1}(1-r^n)}{1-r}$

$S_{10}=\frac{11(1-(\frac{2}{20})^{10})}{1-\frac{2}{20}}$

$S_{10}=\frac{11(1-(\frac{1}{10})^{10}}{\frac{9}{10}} = \frac{1110(1-\frac{1}{10^{10}})}{9}$

$S_{10}=\frac{110}{9}(1-\frac{1}{10^{10}})$

$S_{10}=12.2222...$

Rounding, $S_{10}=12$

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Question

Calculate the sum, rounded to the nearest integer, of the first 16 terms of the following geometric series:

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Answer

This is a geometric series.

The sum of a geometric series can be calculated with the following formula,

$S_{n} = \frac{a_{1}(1-r^n)}{1-r}$ , where n is the number of terms to sum up, r is the common ratio, and $a_{1}$ is the value of the first term.

We have $a_{1}$ and n and we just need to find r before calculating the sum.

Solution:

$a_{1}=6$

$r=\frac{a_{2}}{a_{1}}$

$r=\frac{12.6}{6} = 2.1$

$S_{16}=\frac{a_{1}(1-r^n)}{1-r}$

$S_{16}=\frac{6*(1-2.1^{16})}{1-2.1}$

$S_{16}=780305$

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Question

Use the ratio test to determine if the series diverges or converges:

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

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Answer

$\lim_{n \to \infty}\left | \frac{(n+1)!}{(n+1)^2}\cdot \frac{n^2 } { n!} \right |$

$\lim_{n \to \infty} \left | \frac{(n+1)\cdot n!}{(n+1)(n+1)} \cdot \frac{n^2 }{ n!} \right |$

$\lim_{n \to \infty} \left | \frac{1}{(n+1)} \cdot \frac{n^2 }{1} \right | = \lim_{n \to \infty } \left | \frac{n^2}{n+1}\right |$

This limit is infinite, so the series diverges.

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Question

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

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Answer

$\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ 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 , $\sum_{k=1}^{\infty} \frac{sin(k)}{k^2}$ hence also converges by the comparison test.

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Question

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

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

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Answer

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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Question

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

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Answer

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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Question

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

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Answer

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 . 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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Question

Use the ratio test to determine if this series diverges or converges:

$\sum_{n=0}^{\infty} \frac{n^2}{10^n}$

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Answer

$\lim_{n \to \infty} \left | \frac{(n+1)^2}{10^{n+1}} \cdot \frac{ 10^n}{ n^2 }\right |$

$\lim_{n \to \infty} \left | \frac{(n+1)^2}{10} \cdot \frac{ 1}{ n^2 }\right |$

$\lim_{n \to \infty} \left | \frac{n^2+2n+1}{10n^2} \right | = \frac{1}{10}$

Since the limit is less than 1, the series converges.

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Question

Which of these series cannot be tested for convergence/divergence properly using the ratio test? (Which of these series fails the ratio test?)

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Answer

The ratio test fails when $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | =1$ . Otherwise the series converges absolutely if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | <1$ , and diverges if $\lim_{k\to \infty} \left |\frac{a_{k+1}}{a_k} \right | >1$ .

Testing the series $\sum_{k=0}^{\infty} k$ , we have

$\lim_{k\to\infty} \left |\frac{a_{k+1}}{a_k} \right | = \lim_{k\to\infty} \left |\frac{(k+1)}{(k)} \right | = \lim_{k\to\infty} 1+ \frac{1}{k} = 1.$

Hence the ratio test fails here. (It is likely obvious to the reader that this series diverges already. However, we must remember that all intuition in mathematics requires rigorous justification. We are attempting that here.)

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Question

We consider the following series:

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

Determine the nature of the convergence of the series.

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Answer

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

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Question

We know that :
$chx=\frac{e^x+e^{-x}}{2}$ and $shx=\frac{e^x-e^{-x}}{2}$

We consider the series having the general term:

$v_{n}=\frac{chn}{ch2n}$

Determine the nature of the series:

$\sum_{n=1}^{\infty}v_{n}$

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Answer

We know that:

$chn=\frac{e^n+e^{-n}}{2}$ and therefore we deduce :

$ch2n=\frac{e^{2n }+e^{-2n}}{2}$

We will use the Comparison Test with this problem. To do this we will look at the function in general form $v_{n}\equiv \frac{e^n} {e^{2n}}$ .

We can do this since,

$e^{-n}=\frac{1}{e^n}$ and $e^{-2n}=\frac{1}{e^{2n}}$ approach zero as n approaches infinity. The limit of our function becomes,

$\lim_{n\rightarrow \infty}\frac{e^n+e^{-n}}{e^{2n}+e^{-2n}}=\frac{e^n}{e^{2n}}$

This last part gives us $v_{n}\equiv {e^{-n}}$ .

Now we know that $\frac{1}{e}< 1$ and noting that $e^{-n}}$ is a geometric series that is convergent.

We deduce by the Comparison Test that the series

having general term $v_{n}$ is convergent.

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