Ratio Test and Comparing Series - AP Calculus BC

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Question

Determine what the following series converges to using the Ratio Test, and whether the series is convergent, divergent or neither.

$\sum_{n=0}^{\infty} \frac{(n+1)!(-5)^n}{n!\2^{n+1}}$

Answer

To determine if

$\sum_{n=0}^{\infty} \frac{(n+1)!(-5)^n}{n!\2^{n+1}}$

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,

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

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

$a_n=\frac{(n+1)!(-5)^n}{n!\2^{n+1}}$

and

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

Now

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

We can rearrange the expression to be

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

We can simplify the expression to

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

When we evaluate the limit, we get.

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

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

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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

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

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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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?

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 $\frac{a_{n+1}}{a_n}=\frac{a^{n+1}}{(n+1)^b}\frac{n^b}{a^n}=a\left(\frac{n}{n+1}\right)^b$ .

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

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

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Question

Is the series

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

convergent or divergent, and why?

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.

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Question

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

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

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,

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

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

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

and

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

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

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(-9)^{n+1}}{5^{3n+5}(n+2)}}{\frac{(-9)^n}{5^{3n+2}(n+1)}} \right |$

$=\lim_{n\rightarrow \infty}\left | \frac{(-9)^{n+1}5^{3n+2}(n+1)}{(-9)^n5^{3n+5}(n+2)} \right |$

Now lets simplify this expression to

$=\lim_{n\rightarrow \infty}\left | \frac{(-9)^{1}(n+1)}{5^{3}(n+2)} \right |$

$=\frac{9}{125}\lim_{n\rightarrow \infty}\left | \frac{n+1}{n+2} \right |$

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

Since

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

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

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Question

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

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

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,

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

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.

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Question

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

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

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,

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

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.

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Question

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

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

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,

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

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

$a_{n+1}=\frac{(n+1)!\ 4^{n+1}}{n+2}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)!\ 4^{n+1}}{n+2}}{\frac{n!\ 4^n}{n+1}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)(n+1)!\ 4^{n+1}}{(n+2)n!\ 4^n}\right |$

Now lets simplify this.

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)(n+1) 4}{(n+2)}\right |$

When we evaluate the limit, we get.

$L=\infty$ .

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

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Question

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

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

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,

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

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

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

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.

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Question

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

$\sum_{n=1}^{\infty} \frac{n(-100)^n}{(n+1)5^{n+2}}$

Answer

To determine if

$\sum_{n=1}^{\infty} \frac{n(-100)^n}{(n+1)5^{n+2}}$

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 (therefore convergent).

the series is divergent.

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

Now lets apply this to our situtation.

Let

$a_n=\frac{n(-100)^n}{(n+1)5^{n+2}}$

and

$a_{n+1}=\frac{(n+1)(-100)^{n+1}}{(n+2)5^{n+3}}$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{\frac{(n+1)(-100)^{n+1}}{(n+2)5^{n+3}}}{\frac{n(-100)^n}{(n+1)5^{n+2}}}\right |$

We can rearrange the expression to be

$L=\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2(-100)^{n+1}{5^{n+2}}}{n(n+2)\ (-100)^n \ 5^{n+3}}\right |$

Now lets simplify this.

$L=20\lim_{n\rightarrow \infty}\left | \frac{(n+1)^2}{n(n+2)}\right |$

When we evaluate the limit, we get.

.

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

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Question

Determine the convergence or divergence of the following series:

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

Answer

To determine the convergence or divergence of this series, we use the Ratio Test:

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

If , then the series is absolutely convergent (convergent)

If , then the series is divergent

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

So, we evaluate the limit according to the formula above:

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

which simplified becomes

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

Further simplification results in

$\lim_{n\rightarrow \infty }\left |\frac{2}{n+1} \right |=0<1$

Therefore, the series is absolutely convergent.

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Question

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

$\sum_{n=0}^{\infty} 4^{2n+1}n!$

Answer

To determine if

$\sum_{n=0}^{\infty} 4^{2n+1}n!$

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 thus convergent).

the series is divergent.

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

Now lets apply this to our situtation.

Let

$a_n= 4^{2n+1}n!$

and

$a_{n+1}= 4^{2(n+1)+1}(n+1)!=4^{2n+3}(n+1)!$

Now

$L=\lim_{n\rightarrow \infty}\left | \frac{4^{2n+3}(n+1)!}{4^{2n+1}n!}\right |$

We can simplify the expression to be

$L=\lim_{n\rightarrow \infty}\left | 16 (n+1)}\right |$

When we evaluate the limit, we get.

$L=\infty$ .

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

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