Series of Constants - AP Calculus BC

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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?

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?

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).

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

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

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.

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.

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:

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:

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

Answer

$\lim_{n \to \infty } \left | \frac{(-3)^{n+1}}{(n+1)!} \cdot \frac{ n!}{(-3)^n}\right |$

$\lim_{n \to \infty } \left | \frac{-3 \cdot(-3)^{n}}{(n+1)\cdot n!} \cdot \frac{ n!}{(-3)^n}\right |$

$\lim_{n \to \infty } \left | \frac{-3}{n+1}\right | = 0 < 1$

The series converges.

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Question

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

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

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

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

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

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

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

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

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?)

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.

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:

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

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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Question

Using the Limit Test, determine the nature of the series:

$\frac{n+1}{n^9-7}$

Answer

We will use the Limit Comparison Test to study the nature of the series.

We note first that $n\ge 2$ , the series is positive.

We will compare the general term to $\frac{1}{n^8}$ .

We note that by letting $v_{n}=\frac{n+11}{n^9-7}$ and $u_{n}=\frac{1}{n^8}$ , we have:

$\lim_{n\rightarrow \infty}\frac{u_{n}}{v_n}}=1$ .

Therefore the two series have the same nature, (they either converge or diverge at the same time).

We will use the Integral Test to deduce that the series having the general term:

$u_{n}=\frac{1}{n^8}$ is convergent.

Note that we know that $\int_{1}^{\infty}\frac{1}{x^p}dx$ is convergent if p>1 and in our case p=8 .

This shows that the series having general term $u_{n}$ is convergent.

By the Limit Test, the series having general term is convergent.

This shows that our series is convergent.

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Question

Use the ratio test to determine if the series $\sum_{n=1}^{ \infty } e^{2n+1}$ converges or diverges.

Answer

$\lim_{n \to \infty} \left | \frac{e^{2n+3}}{e^{2n+1}}\right | = \lim_{n \to \infty} \left | e^2 \right | = e^2 > 1$

The series diverges.

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Question

We consider the following series:

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

Determine the nature of the convergence of the series.

Answer

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.

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Question

Using the ratio test,

what can we say about the series.

$\sum_{_n=1}^{\infty} \frac{1}{n^p}$ where is an integer that satisfies:

Answer

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

The Ratio Test states:

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

then if,

L<1 the series converges absolutely.

L>1 the series diverges.

L=1 the series either converges or diverges.

Therefore we need to evaluate,

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

we have,

$a_{n+1}=\frac{1}{(n+1)^p} ,a_{n}=\frac{1}{n^p}$

therefore:

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

We know that

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

and therefore,

$\lim_{n\rightarrow \infty}(\frac{n}{n+1})^p=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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Question

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

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

Answer

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.

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Question

Consider the following series :

$\sum_{n=1}^{\infty}a_{n}$ where $a_{n}$ is given by:

$a_{n}=\frac{n}{n^\alpha +1},\alpha \ge 3$ . Using the ratio test, find the nature of the series.

Answer

Let $a_{n}$ be the general term of the series. We will use the ratio test to check the convergence of the series.

$L=\lim_{n\rightarrow \infty}\left|\frac{a_{n+1}}{a_n}\right|$ if L<1 the series converges absolutely, L>1 the series diverges, and if L=1 the series could either converge or diverge.

We need to evaluate,

$\frac{a_{n+1}}{a_{n}}$ we have:

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

Therefore:

$\frac{a_{n+1}}{a_{n}}=\frac{n^\alpha +1}{(n+1)^\alpha +1}\frac{n+1}{n}$ . We know that,

$\lim_{n\rightarrow \infty}\frac{n}{n+1}=1$ and therefore

$\lim_{n\rightarrow \infty}\frac{n^\alpha +1}{(n+1)^\alpha +1}=1$

This means that :

$\lim_{n\rightarrow \infty}\frac{a_{n}}{a_{n+1}}=1$ .

By the ratio test we can't conclude about the nature of the series. We will have to use another test.

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