Math / Calculus

QuestionConsider the series $\sum_{n=1}^{\infty} \frac{1}{n(n+4)}$
Determine whether the series converges, and if it converges, determine its value. Converges $(\mathrm{y} / \mathrm{n}$ ): $\square$ Value if convergent (blank otherwise): $\square$
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

What is this asking? Does this never-ending sum of fractions, $\sum_{n=1}^{\infty} \frac{1}{n(n+4)}$ , settle down to a specific value, and if so, what is that value? Watch out! Don't forget to check if the series converges before trying to find its value!
A common mistake is to dive right into calculations without ensuring the series actually converges.

STEP 2

1. Check for Convergence
2. Partial Fraction Decomposition
3. Calculate the Telescoping Sum
4. Find the Limit

STEP 3

First, let's see if this series even converges!
We can use the comparison test.
Notice that for every n, $\frac{1}{n(n+4)}$ is smaller than $\frac{1}{n^2}$ .

STEP 4

Why is this helpful?
Because we know that the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges.
It's a famous one, called the p-series with $p=2$ , and p-series converge when $p > 1$ .

STEP 5

Since our series is smaller than a convergent series, our series $\sum_{n=1}^{\infty} \frac{1}{n(n+4)}$ must also converge!
Awesome! Now we can move on to finding its value.

STEP 6

To make this series easier to work with, let's break down each fraction $\frac{1}{n(n+4)}$ into smaller, simpler pieces using partial fraction decomposition.
We want to find constants A and B such that: $\frac{1}{n(n+4)} = \frac{A}{n} + \frac{B}{n+4}$

STEP 7

Multiplying both sides by $n(n+4)$ gives us: $1 = A(n+4) + Bn$ If we let $n = 0$ , we get $1 = 4A$ , so $A = \frac{1}{4}$ .
And if we let $n = -4$ , we get $1 = -4B$ , so $B = -\frac{1}{4}$ .

STEP 8

So, we've found that: $\frac{1}{n(n+4)} = \frac{1/4}{n} - \frac{1/4}{n+4} = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n+4} \right)$ This will make our summation much easier!

STEP 9

Let's write out the first few terms of our series using our new, spiffier fraction form: $\sum_{n=1}^{\infty} \frac{1}{n(n+4)} = \frac{1}{4} \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+4} \right)$

STEP 10

Expanding the first few terms, we get: $\frac{1}{4} \left[ \left( \frac{1}{1} - \frac{1}{5} \right) + \left( \frac{1}{2} - \frac{1}{6} \right) + \left( \frac{1}{3} - \frac{1}{7} \right) + \left( \frac{1}{4} - \frac{1}{8} \right) + \left( \frac{1}{5} - \frac{1}{9} \right) + \dots \right]$

STEP 11

Notice how the $-\frac{1}{5}$ and $\frac{1}{5}$ add to zero!
This is called a telescoping sum.
Many terms will add to zero as we go further down the line.

STEP 12

The terms that don't add to zero are the first four positive terms ( $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{1}{4}$ ) and the "tail end" of negative terms starting with $-\frac{1}{5}$ .

STEP 13

As n approaches infinity, the tail end of negative terms approaches zero.
So, we're left with: $\frac{1}{4} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)$

STEP 14

Calculating this sum: $\frac{1}{4} \left( \frac{12 + 6 + 4 + 3}{12} \right) = \frac{1}{4} \cdot \frac{25}{12} = \frac{25}{48}$

STEP 15

The series converges (y), and its value is $\frac{25}{48}$ .

Was this helpful?

QuestionConsider the series $\sum_{n=1}^{\infty} \frac{1}{n(n+4)}$
Determine whether the series converges, and if it converges, determine its value. Converges $(\mathrm{y} / \mathrm{n}$ ): $\square$ Value if convergent (blank otherwise): $\square$
Note: You can earn partial credit on this problem.

Studdy Solution

STEP 1

STEP 2

1. Check for Convergence
2. Partial Fraction Decomposition
3. Calculate the Telescoping Sum
4. Find the Limit

STEP 3

First, let's see if this series even converges!
We can use the comparison test.
Notice that for every n, $\frac{1}{n(n+4)}$ is smaller than $\frac{1}{n^2}$ .

STEP 4

Why is this helpful?
Because we know that the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges.
It's a famous one, called the p-series with $p=2$ , and p-series converge when $p > 1$ .

STEP 5

Since our series is smaller than a convergent series, our series $\sum_{n=1}^{\infty} \frac{1}{n(n+4)}$ must also converge!
Awesome! Now we can move on to finding its value.

STEP 6

To make this series easier to work with, let's break down each fraction $\frac{1}{n(n+4)}$ into smaller, simpler pieces using partial fraction decomposition.
We want to find constants A and B such that: $\frac{1}{n(n+4)} = \frac{A}{n} + \frac{B}{n+4}$

STEP 7

Multiplying both sides by $n(n+4)$ gives us: $1 = A(n+4) + Bn$ If we let $n = 0$ , we get $1 = 4A$ , so $A = \frac{1}{4}$ .
And if we let $n = -4$ , we get $1 = -4B$ , so $B = -\frac{1}{4}$ .

STEP 8

So, we've found that: $\frac{1}{n(n+4)} = \frac{1/4}{n} - \frac{1/4}{n+4} = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n+4} \right)$ This will make our summation much easier!

STEP 9

Let's write out the first few terms of our series using our new, spiffier fraction form: $\sum_{n=1}^{\infty} \frac{1}{n(n+4)} = \frac{1}{4} \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{n+4} \right)$

STEP 10

Expanding the first few terms, we get: $\frac{1}{4} \left[ \left( \frac{1}{1} - \frac{1}{5} \right) + \left( \frac{1}{2} - \frac{1}{6} \right) + \left( \frac{1}{3} - \frac{1}{7} \right) + \left( \frac{1}{4} - \frac{1}{8} \right) + \left( \frac{1}{5} - \frac{1}{9} \right) + \dots \right]$

STEP 11

Notice how the $-\frac{1}{5}$ and $\frac{1}{5}$ add to zero!
This is called a telescoping sum.
Many terms will add to zero as we go further down the line.

STEP 12

The terms that don't add to zero are the first four positive terms ( $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{1}{4}$ ) and the "tail end" of negative terms starting with $-\frac{1}{5}$ .

STEP 13

As n approaches infinity, the tail end of negative terms approaches zero.
So, we're left with: $\frac{1}{4} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)$

STEP 14

Calculating this sum: $\frac{1}{4} \left( \frac{12 + 6 + 4 + 3}{12} \right) = \frac{1}{4} \cdot \frac{25}{12} = \frac{25}{48}$

STEP 15

The series converges (y), and its value is $\frac{25}{48}$ .

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Studdy Solution

STEP 1

STEP 2

1. Check for Convergence2. Partial Fraction Decomposition3. Calculate the Telescoping Sum4. Find the Limit

STEP 3

First, let's see if this series even converges!We can use the comparison test.Notice that for every *n*, 1n(n+4) \frac{1}{n(n+4)} n(n+4)1​ is *smaller* than 1n2 \frac{1}{n^2} n21​.

STEP 4

Why is this helpful?Because we know that the series ∑n=1∞1n2 \sum_{n=1}^{\infty} \frac{1}{n^2} ∑n=1∞​n21​ converges.It's a famous one, called the **p-series** with p=2p=2p=2, and p-series converge when p>1 p > 1 p>1.

STEP 5

Since our series is *smaller* than a convergent series, our series ∑n=1∞1n(n+4) \sum_{n=1}^{\infty} \frac{1}{n(n+4)} ∑n=1∞​n(n+4)1​ must also **converge**!Awesome! Now we can move on to finding its value.

STEP 6

STEP 7

Multiplying both sides by n(n+4) n(n+4) n(n+4) gives us: 1=A(n+4)+Bn 1 = A(n+4) + Bn 1=A(n+4)+Bn If we let n=0 n = 0 n=0, we get 1=4A 1 = 4A 1=4A, so A=14 A = \frac{1}{4} A=41​.And if we let n=−4 n = -4 n=−4, we get 1=−4B 1 = -4B 1=−4B, so B=−14 B = -\frac{1}{4} B=−41​.

STEP 8

So, we've found that: 1n(n+4)=1/4n−1/4n+4=14(1n−1n+4) \frac{1}{n(n+4)} = \frac{1/4}{n} - \frac{1/4}{n+4} = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n+4} \right) n(n+4)1​=n1/4​−n+41/4​=41​(n1​−n+41​) This will make our summation *much* easier!

STEP 9

STEP 10

STEP 11

Notice how the −15 -\frac{1}{5} −51​ and 15 \frac{1}{5} 51​ add to zero!This is called a **telescoping sum**.Many terms will add to zero as we go further down the line.

STEP 12

The terms that *don't* add to zero are the first four positive terms (11 \frac{1}{1} 11​, 12 \frac{1}{2} 21​, 13 \frac{1}{3} 31​, 14 \frac{1}{4} 41​) and the "tail end" of negative terms starting with −15 -\frac{1}{5} −51​.

STEP 13

As *n* approaches infinity, the tail end of negative terms approaches zero.So, we're left with: 14(11+12+13+14) \frac{1}{4} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) 41​(11​+21​+31​+41​)

STEP 14

Calculating this sum: 14(12+6+4+312)=14⋅2512=2548 \frac{1}{4} \left( \frac{12 + 6 + 4 + 3}{12} \right) = \frac{1}{4} \cdot \frac{25}{12} = \frac{25}{48} 41​(1212+6+4+3​)=41​⋅1225​=4825​

STEP 15

The series converges (y), and its value is 2548 \frac{25}{48} 4825​.

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1. Check for Convergence
2. Partial Fraction Decomposition
3. Calculate the Telescoping Sum
4. Find the Limit

First, let's see if this series even converges!
We can use the comparison test.
Notice that for every n, $\frac{1}{n(n+4)}$ is smaller than $\frac{1}{n^2}$ .

Why is this helpful?
Because we know that the series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges.
It's a famous one, called the p-series with $p=2$ , and p-series converge when $p > 1$ .

Since our series is smaller than a convergent series, our series $\sum_{n=1}^{\infty} \frac{1}{n(n+4)}$ must also converge!
Awesome! Now we can move on to finding its value.

Multiplying both sides by $n(n+4)$ gives us: $1 = A(n+4) + Bn$ If we let $n = 0$ , we get $1 = 4A$ , so $A = \frac{1}{4}$ .
And if we let $n = -4$ , we get $1 = -4B$ , so $B = -\frac{1}{4}$ .

So, we've found that: $\frac{1}{n(n+4)} = \frac{1/4}{n} - \frac{1/4}{n+4} = \frac{1}{4} \left( \frac{1}{n} - \frac{1}{n+4} \right)$ This will make our summation much easier!

Notice how the $-\frac{1}{5}$ and $\frac{1}{5}$ add to zero!
This is called a telescoping sum.
Many terms will add to zero as we go further down the line.

The terms that don't add to zero are the first four positive terms ( $\frac{1}{1}$ , $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{1}{4}$ ) and the "tail end" of negative terms starting with $-\frac{1}{5}$ .

As n approaches infinity, the tail end of negative terms approaches zero.
So, we're left with: $\frac{1}{4} \left( \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)$

Calculating this sum: $\frac{1}{4} \left( \frac{12 + 6 + 4 + 3}{12} \right) = \frac{1}{4} \cdot \frac{25}{12} = \frac{25}{48}$

The series converges (y), and its value is $\frac{25}{48}$ .