Series: Videos & Practice Problems

To find the first four terms of the sequence defined by $a_n=n^2n_{-}!$, substitute values of $n$ from 1 to 4. The calculations yield: $a_1=1$, $a_2=4$, $a_3=18$, and $a_4=96$. Understanding factorials and their application in sequences is crucial for further mathematical concepts.

Intro to Series: Partial Sums

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

Knowing that \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Since \( 2! = 2 \), we calculate:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Knowing that \( 3! = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Intro to Series: Partial Sums Example 1

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Compute the first four partial sums and find a formula for the $n^{\mathrm{th}}$ partial sum.
${\sum }_{n=1}^{\infty }2n-1$

• A. $1,3,5,9;S_n=2n-1$
• B. $1,4,9,16;S_n=n^2$
• C. $1,4,9,16;S_n=2n-1$
• D. $1,3,5,9;S_n=n^2$

Answer: (B)

Convergence of an Infinite Series

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

Knowing that \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Since \( 2! = 2 \), we calculate:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Knowing that \( 3! = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Convergence of an Infinite Series Example 2

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Geometric Series

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

And since \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Calculating \( 2! = 2 \times 1 = 2 \), we get:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Since \( 3! = 3 \times 2 \times 1 = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Geometric Series Example 3

To find the first four terms of the sequence defined by \( a_n = n^2 \times (n-1)! \), we will evaluate the expression for \( n = 1 \) through \( n = 4 \).

Starting with \( n = 1 \):

We have:

\( a_1 = 1^2 \times (1-1)! \)

Calculating this gives:

\( a_1 = 1 \times 0! \)

Since \( 0! = 1 \), we find:

\( a_1 = 1 \times 1 = 1 \)

Next, for \( n = 2 \):

We calculate:

\( a_2 = 2^2 \times (2-1)! \)

This simplifies to:

\( a_2 = 4 \times 1! \)

Knowing that \( 1! = 1 \), we have:

\( a_2 = 4 \times 1 = 4 \)

Now, for \( n = 3 \):

We find:

\( a_3 = 3^2 \times (3-1)! \)

This becomes:

\( a_3 = 9 \times 2! \)

Since \( 2! = 2 \), we calculate:

\( a_3 = 9 \times 2 = 18 \)

Finally, for \( n = 4 \):

We compute:

\( a_4 = 4^2 \times (4-1)! \)

This simplifies to:

\( a_4 = 16 \times 3! \)

Knowing that \( 3! = 6 \), we find:

\( a_4 = 16 \times 6 = 96 \)

Thus, the first four terms of the sequence are \( 1, 4, 18, \) and \( 96 \). This sequence can be extended further by continuing to apply the same formula for higher values of \( n \).

Here’s what students ask on this topic:

How do you find the first four terms of a sequence defined by a_n = n² × (n-1)!?

To find the first four terms of the sequence a_n = n² × (n-1)!, substitute n = 1, 2, 3, and 4 into the formula:

For n = 1: a₁ = 1² × 0! = 1 × 1 = 1.

For n = 2: a₂ = 2² × 1! = 4 × 1 = 4.

For n = 3: a₃ = 3² × 2! = 9 × (2 × 1) = 18.

For n = 4: a₄ = 4² × 3! = 16 × (3 × 2 × 1) = 96.

Thus, the first four terms are 1, 4, 18, and 96.

What is the significance of factorials in sequences?

Factorials, denoted as n!, are significant in sequences because they represent the product of all positive integers up to n. They are commonly used in combinatorics, probability, and series calculations. In sequences, factorials can introduce rapid growth, as each term depends on the multiplication of increasingly larger numbers. For example, in the sequence a_n = n² × (n-1)!, the factorial component (n-1)! causes the terms to grow much faster than a simple polynomial sequence. Understanding factorials is essential for solving problems involving permutations, combinations, and series expansions.

How do you simplify expressions involving factorials?

To simplify expressions involving factorials, use the definition of factorial: n! = n × (n-1) × (n-2) × ... × 1. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. Factorials can also be simplified by canceling terms when dividing. For instance, $\frac{\left(5\right)!}{\left(3\right)!}$ simplifies to 5 × 4 = 20 because the 3! terms cancel out. Additionally, remember that 0! = 1 by definition. Simplifying factorials is crucial for solving series and combinatorial problems efficiently.

What are common applications of sequences in calculus?

Sequences are widely used in calculus for analyzing patterns, approximating functions, and solving problems involving limits. They form the foundation for series, which are sums of sequence terms. Applications include Taylor and Maclaurin series for function approximation, convergence tests for infinite series, and modeling real-world phenomena like population growth or financial investments. Understanding sequences helps in studying the behavior of functions as n approaches infinity, which is essential for solving problems in advanced calculus and mathematical analysis.

How do you determine if a sequence converges or diverges?

To determine if a sequence converges or diverges, analyze its limit as n approaches infinity. If the terms of the sequence approach a finite value, the sequence converges. For example, the sequence a_n = 1/n converges to 0 as n → ∞. If the terms grow without bound or oscillate indefinitely, the sequence diverges. For instance, a_n = n diverges because its terms increase infinitely. Techniques like the squeeze theorem or direct computation of limits are often used to assess convergence or divergence.