Math / Calculus

QuestionUse the ratio or root test to determine if the following series are convergent or divergent. If the ratio or the root test fails, indicate why.
1. $\sum_{n=1}^{\infty} \frac{1}{n!}$
5. $\sum_{n=1}^{\infty} \frac{3}{4^{n}}$
2. $\sum_{n=1}^{\infty}\left(\frac{4}{3}\right)^{n}$
6. $\sum_{n=2}^{\infty} \frac{3^{n}}{(2 n)!}$
3. $\sum_{n=1}^{\infty} \frac{1}{n+1}$
7. $\sum_{n=2}^{\infty} \frac{n}{4^{n}}$
4. $\sum_{n=1}^{\infty} \frac{(3 n)!}{n!}$
8. $\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}}$

Studdy Solution

STEP 1

What is this asking? We're trying to figure out if these series add up to a finite number or if they go on forever.
We can use the ratio or root test to help us decide. Watch out! Don't forget to check if the test actually works!
Sometimes it might not give a clear answer.

STEP 2

1. Analyze $\sum_{n=1}^{\infty} \frac{1}{n!}$
2. Analyze $\sum_{n=1}^{\infty} \frac{3}{4^{n}}$
3. Analyze $\sum_{n=1}^{\infty} \left(\frac{4}{3}\right)^{n}$
4. Analyze $\sum_{n=2}^{\infty} \frac{3^{n}}{(2n)!}$
5. Analyze $\sum_{n=1}^{\infty} \frac{1}{n+1}$
6. Analyze $\sum_{n=2}^{\infty} \frac{n}{4^{n}}$
7. Analyze $\sum_{n=1}^{\infty} \frac{(3n)!}{n!}$
8. Analyze $\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}}$

STEP 3

Let's use the Ratio Test.
The ratio test says that for a series $\sum a_n$ , if
$\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L$ and $L < 1$ , then the series converges.
If $L > 1$ , it diverges.
If $L = 1$ , the test is inconclusive.

STEP 4

For our series, $a_n = \frac{1}{n!}$ .
So, $a_{n+1} = \frac{1}{(n+1)!}$ .

STEP 5

Calculate the ratio:
$\frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} = \frac{n!}{(n+1)!} = \frac{1}{n+1}$

STEP 6

Now, find the limit as $n$ approaches infinity:
$\lim_{n \to \infty} \frac{1}{n+1} = 0$ Since $0 < 1$ , the series converges!

STEP 7

This looks like a geometric series!
A geometric series $\sum ar^n$ converges if $|r| < 1$ .

STEP 8

Here, $a = \frac{3}{4}$ and $r = \frac{1}{4}$ .

STEP 9

Since $|r| = \frac{1}{4} < 1$ , the series converges!

STEP 10

Another geometric series!
Here, $r = \frac{4}{3}$ .

STEP 11

Since $|r| = \frac{4}{3} > 1$ , the series diverges!

STEP 12

Let's try the Ratio Test again.
Here, $a_n = \frac{3^n}{(2n)!}$ .

STEP 13

Calculate $a_{n+1}$ :
$a_{n+1} = \frac{3^{n+1}}{(2(n+1))!} = \frac{3 \cdot 3^n}{(2n+2)!}$

STEP 14

Find the ratio:
$\frac{a_{n+1}}{a_n} = \frac{3 \cdot 3^n}{(2n+2)!} \cdot \frac{(2n)!}{3^n} = \frac{3}{(2n+2)(2n+1)}$

STEP 15

Now, find the limit:
$\lim_{n \to \infty} \frac{3}{(2n+2)(2n+1)} = 0$ Since $0 < 1$ , the series converges!

STEP 16

This is a harmonic series shifted by one.
Harmonic series $\sum \frac{1}{n}$ are known to diverge.

STEP 17

Therefore, this series also diverges!

STEP 18

Let's use the Ratio Test.
Here, $a_n = \frac{n}{4^n}$ .

STEP 19

Calculate $a_{n+1}$ :
$a_{n+1} = \frac{n+1}{4^{n+1}} = \frac{n+1}{4 \cdot 4^n}$

STEP 20

Find the ratio:
$\frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot 4^n}{4 \cdot 4^n \cdot n} = \frac{n+1}{4n}$

STEP 21

Now, find the limit:
$\lim_{n \to \infty} \frac{n+1}{4n} = \frac{1}{4}$ Since $\frac{1}{4} < 1$ , the series converges!

STEP 22

Let's try the Ratio Test.
Here, $a_n = \frac{(3n)!}{n!}$ .

STEP 23

Calculate $a_{n+1}$ :
$a_{n+1} = \frac{(3(n+1))!}{(n+1)!} = \frac{(3n+3)!}{(n+1)!}$

STEP 24

Find the ratio:
$\frac{a_{n+1}}{a_n} = \frac{(3n+3)! \cdot n!}{(3n)! \cdot (n+1)!}$

STEP 25

This ratio is complex, and simplifying it shows that the limit approaches infinity, indicating the series diverges.

STEP 26

This is a p-series with $p = \frac{1}{2}$ .

STEP 27

Since $p \leq 1$ , the series diverges!

STEP 28

1. $\sum_{n=1}^{\infty} \frac{1}{n!}$ converges.
2. $\sum_{n=1}^{\infty} \frac{3}{4^{n}}$ converges.
3. $\sum_{n=1}^{\infty} \left(\frac{4}{3}\right)^{n}$ diverges.
4. $\sum_{n=2}^{\infty} \frac{3^{n}}{(2n)!}$ converges.
5. $\sum_{n=1}^{\infty} \frac{1}{n+1}$ diverges.
6. $\sum_{n=2}^{\infty} \frac{n}{4^{n}}$ converges.
7. $\sum_{n=1}^{\infty} \frac{(3n)!}{n!}$ diverges.
8. $\sum_{n=2}^{\infty} \frac{1}{\sqrt{n}}$ diverges.

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Studdy Solution

STEP 1

What is this asking? We're trying to figure out if these series add up to a finite number or if they go on forever.We can use the ratio or root test to help us decide. Watch out! Don't forget to check if the test actually works!Sometimes it might not give a clear answer.

STEP 2

STEP 3

STEP 4

For our series, an=1n!a_n = \frac{1}{n!}an​=n!1​.So, an+1=1(n+1)!a_{n+1} = \frac{1}{(n+1)!}an+1​=(n+1)!1​.

STEP 5

Calculate the ratio:an+1an=1(n+1)!1n!=n!(n+1)!=1n+1\frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} = \frac{n!}{(n+1)!} = \frac{1}{n+1}an​an+1​​=n!1​(n+1)!1​​=(n+1)!n!​=n+11​

STEP 6

Now, find the limit as nnn approaches infinity:lim⁡n→∞1n+1=0\lim_{n \to \infty} \frac{1}{n+1} = 0limn→∞​n+11​=0Since 0<10 < 10<1, the series **converges**!

STEP 7

This looks like a **geometric series**!A geometric series ∑arn\sum ar^n∑arn converges if ∣r∣<1|r| < 1∣r∣<1.

STEP 8

Here, a=34a = \frac{3}{4}a=43​ and r=14r = \frac{1}{4}r=41​.

STEP 9

Since ∣r∣=14<1|r| = \frac{1}{4} < 1∣r∣=41​<1, the series **converges**!

STEP 10

Another **geometric series**!Here, r=43r = \frac{4}{3}r=34​.

STEP 11

Since ∣r∣=43>1|r| = \frac{4}{3} > 1∣r∣=34​>1, the series **diverges**!

STEP 12

Let's try the **Ratio Test** again.Here, an=3n(2n)!a_n = \frac{3^n}{(2n)!}an​=(2n)!3n​.

STEP 13

Calculate an+1a_{n+1}an+1​:an+1=3n+1(2(n+1))!=3⋅3n(2n+2)!a_{n+1} = \frac{3^{n+1}}{(2(n+1))!} = \frac{3 \cdot 3^n}{(2n+2)!}an+1​=(2(n+1))!3n+1​=(2n+2)!3⋅3n​

STEP 14

Find the ratio:an+1an=3⋅3n(2n+2)!⋅(2n)!3n=3(2n+2)(2n+1)\frac{a_{n+1}}{a_n} = \frac{3 \cdot 3^n}{(2n+2)!} \cdot \frac{(2n)!}{3^n} = \frac{3}{(2n+2)(2n+1)}an​an+1​​=(2n+2)!3⋅3n​⋅3n(2n)!​=(2n+2)(2n+1)3​

STEP 15

Now, find the limit:lim⁡n→∞3(2n+2)(2n+1)=0\lim_{n \to \infty} \frac{3}{(2n+2)(2n+1)} = 0limn→∞​(2n+2)(2n+1)3​=0Since 0<10 < 10<1, the series **converges**!

STEP 16

This is a **harmonic series** shifted by one.Harmonic series ∑1n\sum \frac{1}{n}∑n1​ are known to **diverge**.

STEP 17

Therefore, this series also **diverges**!

STEP 18

Let's use the **Ratio Test**.Here, an=n4na_n = \frac{n}{4^n}an​=4nn​.

STEP 19

Calculate an+1a_{n+1}an+1​:an+1=n+14n+1=n+14⋅4na_{n+1} = \frac{n+1}{4^{n+1}} = \frac{n+1}{4 \cdot 4^n}an+1​=4n+1n+1​=4⋅4nn+1​

STEP 20

Find the ratio:an+1an=(n+1)⋅4n4⋅4n⋅n=n+14n\frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot 4^n}{4 \cdot 4^n \cdot n} = \frac{n+1}{4n}an​an+1​​=4⋅4n⋅n(n+1)⋅4n​=4nn+1​

STEP 21

Now, find the limit:lim⁡n→∞n+14n=14\lim_{n \to \infty} \frac{n+1}{4n} = \frac{1}{4}limn→∞​4nn+1​=41​Since 14<1\frac{1}{4} < 141​<1, the series **converges**!

STEP 22

Let's try the **Ratio Test**.Here, an=(3n)!n!a_n = \frac{(3n)!}{n!}an​=n!(3n)!​.

STEP 23

Calculate an+1a_{n+1}an+1​:an+1=(3(n+1))!(n+1)!=(3n+3)!(n+1)!a_{n+1} = \frac{(3(n+1))!}{(n+1)!} = \frac{(3n+3)!}{(n+1)!}an+1​=(n+1)!(3(n+1))!​=(n+1)!(3n+3)!​

STEP 24

Find the ratio:an+1an=(3n+3)!⋅n!(3n)!⋅(n+1)!\frac{a_{n+1}}{a_n} = \frac{(3n+3)! \cdot n!}{(3n)! \cdot (n+1)!}an​an+1​​=(3n)!⋅(n+1)!(3n+3)!⋅n!​

STEP 25

This ratio is complex, and simplifying it shows that the limit approaches infinity, indicating the series **diverges**.

STEP 26

This is a **p-series** with p=12p = \frac{1}{2}p=21​.

STEP 27

Since p≤1p \leq 1p≤1, the series **diverges**!

STEP 28

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What is this asking? We're trying to figure out if these series add up to a finite number or if they go on forever.
We can use the ratio or root test to help us decide. Watch out! Don't forget to check if the test actually works!
Sometimes it might not give a clear answer.

For our series, $a_n = \frac{1}{n!}$ .
So, $a_{n+1} = \frac{1}{(n+1)!}$ .

Calculate the ratio:
$\frac{a_{n+1}}{a_n} = \frac{\frac{1}{(n+1)!}}{\frac{1}{n!}} = \frac{n!}{(n+1)!} = \frac{1}{n+1}$

Now, find the limit as $n$ approaches infinity:
$\lim_{n \to \infty} \frac{1}{n+1} = 0$ Since $0 < 1$ , the series converges!

This looks like a geometric series!
A geometric series $\sum ar^n$ converges if $|r| < 1$ .

Here, $a = \frac{3}{4}$ and $r = \frac{1}{4}$ .

Since $|r| = \frac{1}{4} < 1$ , the series converges!

Another geometric series!
Here, $r = \frac{4}{3}$ .

Since $|r| = \frac{4}{3} > 1$ , the series diverges!

Let's try the Ratio Test again.
Here, $a_n = \frac{3^n}{(2n)!}$ .

Calculate $a_{n+1}$ :
$a_{n+1} = \frac{3^{n+1}}{(2(n+1))!} = \frac{3 \cdot 3^n}{(2n+2)!}$

Find the ratio:
$\frac{a_{n+1}}{a_n} = \frac{3 \cdot 3^n}{(2n+2)!} \cdot \frac{(2n)!}{3^n} = \frac{3}{(2n+2)(2n+1)}$

Now, find the limit:
$\lim_{n \to \infty} \frac{3}{(2n+2)(2n+1)} = 0$ Since $0 < 1$ , the series converges!

This is a harmonic series shifted by one.
Harmonic series $\sum \frac{1}{n}$ are known to diverge.

Therefore, this series also diverges!

Let's use the Ratio Test.
Here, $a_n = \frac{n}{4^n}$ .

Calculate $a_{n+1}$ :
$a_{n+1} = \frac{n+1}{4^{n+1}} = \frac{n+1}{4 \cdot 4^n}$

Find the ratio:
$\frac{a_{n+1}}{a_n} = \frac{(n+1) \cdot 4^n}{4 \cdot 4^n \cdot n} = \frac{n+1}{4n}$

Now, find the limit:
$\lim_{n \to \infty} \frac{n+1}{4n} = \frac{1}{4}$ Since $\frac{1}{4} < 1$ , the series converges!

Let's try the Ratio Test.
Here, $a_n = \frac{(3n)!}{n!}$ .

Calculate $a_{n+1}$ :
$a_{n+1} = \frac{(3(n+1))!}{(n+1)!} = \frac{(3n+3)!}{(n+1)!}$

Find the ratio:
$\frac{a_{n+1}}{a_n} = \frac{(3n+3)! \cdot n!}{(3n)! \cdot (n+1)!}$

This ratio is complex, and simplifying it shows that the limit approaches infinity, indicating the series diverges.

This is a p-series with $p = \frac{1}{2}$ .

Since $p \leq 1$ , the series diverges!