Ratio Test and Comparing Series - AP Calculus BC

Card 1 of 280

0

Didn't Know

0

Didn't Know

Knew It

0

1 of 2019 left

Question

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

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

Tap to reveal answer

Answer

The series converges.

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

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,

.

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

Now

We can rearrange the expression to be

We can simplify the expression to

When we evaluate the limit, we get.

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

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

This limit is infinite, so the series diverges.

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

The ratio test fails when . Otherwise the series converges absolutely if , and diverges if .

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

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.)

← Didn't Know|Knew It →

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.

Tap to reveal answer

Answer

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

is positive.

We have all natural numbers n:

, this implies that

.

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

← Didn't Know|Knew It →

Question

We know that :
and

We consider the series having the general term:

Determine the nature of the series:

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

Tap to reveal answer

Answer

We know that:

and therefore we deduce :

We will use the Comparison Test with this problem. To do this we will look at the function in general form .

We can do this since,

and approach zero as n approaches infinity. The limit of our function becomes,

This last part gives us .

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

We deduce by the Comparison Test that the series

having general term is convergent.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

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 .

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

.

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 is convergent.

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

This shows that our series is convergent.

← Didn't Know|Knew It →

Question

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

Tap to reveal answer

Answer

The series diverges.

← Didn't Know|Knew It →

Question

We consider the following series:

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

Determine the nature of the convergence of the series.

Tap to reveal answer

Answer

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

is positive.

We have all natural numbers n:

, this implies that

.

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.

← Didn't Know|Knew It →

Question

Using the ratio test,

what can we say about the series.

where is an integer that satisfies:

Tap to reveal answer

Answer

Let be the general term of the series. We will use the ratio test to check the convergence of the series.

The Ratio Test states:

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,

we have,

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

therefore:

.

We know that

and therefore,

This means that :

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

← Didn't Know|Knew It →

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

Tap to reveal answer

Answer

As required by this question we will have to use the ratio test. 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 : . In our case:

Therefore

.

We know that

This means that

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

← Didn't Know|Knew It →

Question

Consider the following series :

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

. Using the ratio test, find the nature of the series.

Tap to reveal answer

Answer

Let be the general term of the series. We will use the ratio test to check the convergence of the series.

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,

we have:

.

Therefore:

. We know that,

and therefore

This means that :

.

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

← Didn't Know|Knew It →

Question

We consider the series,

$\sum_${n=1}^{\infty}$$$\frac{a^n$$}{n^b$}$,0<a<$\frac{1}{999}$ ,b>0$ .

Using the ratio test, what can we conclude about the nature of convergence of this series?

Tap to reveal answer

Answer

Note that the series is positive.

As it is required we will use the ratio test to check for the nature of the series.

We have .

Therefore, $\lim_${n\rightarrow\infty}$\frac{n^b$$}{(n+1)^b$}$=a(\lim_${n\rightarrow}$\frac{n}{n+1}$)^b$=a$

if L>1 the series diverges, if L<1 the series converges absolutely, and if L=1 the series may either converge or diverge.

Since $a<$\frac{1}{999}$<1$ the ratio test concludes that the series converges absolutely.

← Didn't Know|Knew It →

Question

Is the series

$\small $\sum_{n=1}$^\infty $$\frac{n^3$$}{n^7$+1000}$$

convergent or divergent, and why?

Tap to reveal answer

Answer

We will use the comparison test to prove that

$\small $\sum_{n=1}$^\infty $$\frac{n^3$$}{n^7$+1000}$$

converges (Note: we cannot use the ratio test, because then the ratio will be $\small 1$ , which means the test is inconclusive).

We will compare $\small $$\frac{n^3$$}{n^7$+1000}$$ to $\small $$\frac{1}{n^4$}$$ because they "behave" somewhat similarly. Both series are nonzero for all $\small n\geq 1$ , so one of the conditions is satisfied.

The series

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

converges, so we must show that

$\small \small $$\frac{1}{n^4$}$\geq $$\frac{n^3$$}{n^4$+1000}$$

for $\small n\geq 1$ .

This is easy to show because

$\small $$\frac{1}{n^4$$}$=\frac{n^3$$}{n^7$}$\geq $$\frac{n^3$$}{n^7$+1000}$$

since the denominator $\small $n^7$+1000$ is greater than or equal to $\small $n^7$$ for all $\small n\geq 1$ .

Thus, since

$\small \small $$\frac{1}{n^4$}$\geq $$\frac{n^3$$}{n^4$+1000}$$

and because

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

converges, it follows that

$\small $\sum_{n=1}$^\infty $$\frac{n^3$$}{n^7$+1000}$$

converges, by comparison test.

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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,

Then if

, the series is absolutely convergent.

, the series is divergent.

, 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}$((n+1)+1)}$

Now

Now lets simplify this expression to

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

Since

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

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

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,

Then if

, the series is absolutely convergent.

, the series is divergent.

, 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

.

Now lets simplify this expression to

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

Since $\infty>1$ ,

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

← Didn't Know|Knew It →

Question

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

$\sum_${n=1}^{\infty}$ $$\frac{11^{n}$$$}{(n+1)3^{n}$}$

Tap to reveal answer

Answer

In order to figure if

$\sum_${n=1}^{\infty}$ $$\frac{11^{n}$$$}{(n+1)3^{n}$}$

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,

Then if

, the series is absolutely convergent.

, the series is divergent.

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

Now lets apply the ratio test to our problem.

Let

$a_$n=\frac{11^n$$}{(n+1)3^{n}$$}$

and

$a_${n+1}=\frac{11^{n+1}$$$}{(n+2)3^{n+1}$}$ .

Now

.

Now lets simplify this expression to

$=\frac{11}{3}$*1=\frac{11}{3}$$ .

Since $$\frac{11}{3}$>1$ ,

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

← Didn't Know|Knew It →

Question

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

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

Tap to reveal answer

Answer

To determine if

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

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,

.

If

the series is absolutely convergent (and therefore 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!\ $4^n$}{n+1}$$

and

Now

We can rearrange the expression to be

Now lets simplify this.

When we evaluate the limit, we get.

$L=\infty$ .

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

← Didn't Know|Knew It →

Question

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

$\sum_${n=0}^{\infty}$ $\frac{n \ $2^n$$}{5^n$}$$

Tap to reveal answer

Answer

To determine if

$\sum_${n=0}^{\infty}$ $\frac{n \ $2^n$$}{5^n$}$$

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,

.

If

the series is absolutely convergent (and thus 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^n$$}{5^n$}$$

and

Now

We can rearrange the expression to be

.

Now lets simplify this.

When we evaluate the limit, we get.

$L=\frac{2}{5}$$ .

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

← Didn't Know|Knew It →

0

Knew It