AP Calculus BC - Ratio Test and Comparing Series

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

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

We cannot conclude when we use the ratio test.

It is convergent.

$\infty$

$-\infty$

Explanation

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.

We consider the following series:

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

Determine the nature of the convergence of the series.

The series is divergent.

$\frac{1}{2}$

$\frac{3}4{}$

Explanation

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

We consider the following series:

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

Determine the nature of the convergence of the series.

The series is divergent.

$\frac{1}{2}$

$\frac{3}4{}$

Explanation

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.

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

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

Convegent

Divergent

Neither

More tests are needed.

Inconclusive

Explanation

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,

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

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

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

Now

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

We can rearrange the expression to be

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

Now lets simplify this.

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

When we evaluate the limit, we get.

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

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

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:

We can't conclude when we use the ratio test.

$\infty$

We can't use the ratio test to study this series.

$\frac{4}{5}$

Explanation

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.

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

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

The series converges.

The series diverges.

Unable to determine

Explanation

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

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

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

The series converges.

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

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

Divergent

Neither

Convergent

Both

Inconclusive

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

, 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

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

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

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.

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

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

The series converges

The series diverges

Unable to determine

Explanation

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

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

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

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

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

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

Divergent

Convergent

Neither

Both

Inconclusive

Explanation

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,

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

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

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{11^{n+1}}{(n+2)3^{n+1}}}{\frac{11^n}{(n+1)3^{n}}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{(n+1)11^{n+1}3^n}{11^{n}(n+2)3^{n+1}} \right |$ .

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{11(n+1)}{3(n+2)} \right |$

$=\frac{11}{3}\lim_{n\rightarrow \infty}\left |\frac{(n+1)}{(n+2)} \right |$

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

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

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

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

The series diverges.

The series converges.

Unable to determine

Explanation

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

The series diverges.

Ratio Test and Comparing Series

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