Series: Videos & Practice Problems

Understanding sequences is fundamental in mathematics, and transitioning from sequences to partial sums is a crucial step in grasping the concept of series. A sequence is essentially a list of numbers that follow a specific pattern, while a partial sum refers to the sum of the first n terms of a sequence, denoted as S_n.

To illustrate this, consider the sequence defined by a_n = \frac{1}{2^n}. The first five terms of this sequence can be calculated as follows:

• a₁ = \frac{1}{2^1} = \frac{1}{2}
• a₂ = \frac{1}{2^2} = \frac{1}{4}
• a₃ = \frac{1}{2^3} = \frac{1}{8}
• a₄ = \frac{1}{2^4} = \frac{1}{16}
• a₅ = \frac{1}{2^5} = \frac{1}{32}

Next, we can find the first five partial sums:

• S₁ = a₁ = \frac{1}{2}
• S₂ = a₁ + a₂ = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}
• S₃ = a₁ + a₂ + a₃ = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{7}{8}
• S₄ = a₁ + a₂ + a₃ + a₄ = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} = \frac{15}{16}
• S₅ = a₁ + a₂ + a₃ + a₄ + a₅ = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \frac{1}{32} = \frac{31}{32}

To derive the nth partial sum, we observe the pattern in the denominators and numerators of the fractions. The denominators are powers of 2: 2¹, 2², 2³, ... , 2ⁿ. The numerators are always one less than the corresponding denominator, leading to the formula for the nth partial sum:

S_n = \frac{2^n - 1}{2^n}

This relationship shows that as n increases, the partial sums approach 1, illustrating the concept of convergence in series. Additionally, the collection of partial sums S₁, S₂, S₃, ... , S_n can be viewed as a new sequence, further emphasizing the interconnectedness of these mathematical concepts.

To solidify your understanding, practicing with various sequences and their corresponding partial sums will be beneficial. Engaging with exercises will enhance your mastery of sequences, partial sums, and the broader concept of series.

To find the formula for the nth partial sum \( S_n \) of the sequence defined by \( a_n = \frac{1}{n} - \frac{1}{n+1} \), we start by calculating the first few terms of the series. The first term, \( S_1 \), is simply the first term of the sequence:

\( S_1 = a_1 = \frac{1}{1} - \frac{1}{2} = \frac{1}{2} \).

Next, for \( S_2 \), we add the second term \( a_2 \) to \( S_1 \):

\( S_2 = S_1 + a_2 = \frac{1}{2} + \left( \frac{1}{2} - \frac{1}{3} \right) = \frac{1}{2} + \frac{1}{2} - \frac{1}{3} = 1 - \frac{1}{3} = \frac{2}{3} \).

Continuing this process, we find \( S_3 \):

\( S_3 = S_2 + a_3 = \frac{2}{3} + \left( \frac{1}{3} - \frac{1}{4} \right) = \frac{2}{3} + \frac{1}{3} - \frac{1}{4} = 1 - \frac{1}{4} = \frac{3}{4} \).

As we calculate more terms, a pattern emerges. Each partial sum \( S_n \) can be expressed as:

\( S_n = 1 - \frac{1}{n+1} \).

This formula indicates that the first term is always 1, and the last term is the negative fraction \( -\frac{1}{n+1} \). The cancellation of intermediate terms leads to this simplified expression.

To find the sum of the first 15 terms, we substitute \( n = 15 \) into our formula:

\( S_{15} = 1 - \frac{1}{15 + 1} = 1 - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16} \).

This result shows that the sum of the first 15 terms of the sequence is \( \frac{15}{16} \). By deriving the nth partial sum formula, we can efficiently calculate the sum for any number of terms without needing to compute each term individually.

In the study of infinite series, we explore the sum of infinitely many terms in a sequence. Unlike finite series, where we add a specific number of terms, infinite series involve evaluating the limit of the sum as the number of terms approaches infinity. This limit, denoted as \( s_n \), is crucial in determining whether the series converges or diverges. If the limit exists and is a finite value, we say the series converges. Conversely, if the limit does not exist or approaches infinity, the series diverges.

To illustrate this concept, consider a series where we identify a pattern by writing out the first few terms. For example, if we have a series defined by terms like \( \frac{1}{n+1} - \frac{1}{n+2} \), we can compute the first few sums, \( s_1, s_2, \) and \( s_3 \). As we calculate these sums, we notice that many terms cancel each other out, leading to a simplified expression. This cancellation helps us derive a general formula for the \( n \)-th term of the series.

For instance, if we find that the \( n \)-th term can be expressed as \( s_n = \frac{1}{2} - \frac{1}{n+2} \), we can then evaluate the limit as \( n \) approaches infinity. In this case, the limit simplifies to \( \frac{1}{2} - 0 = \frac{1}{2} \), indicating that the series converges to \( \frac{1}{2} \).

Additionally, it is important to recognize that if we have two convergent series, \( a_n \) and \( b_n \), their sum or difference will also converge. Furthermore, multiplying a convergent series by a constant will maintain convergence. This principle allows for flexibility in working with infinite series, providing a robust framework for analysis.

In summary, understanding the behavior of infinite series through limits and recognizing patterns in terms can greatly aid in determining convergence. By applying these concepts, we can effectively analyze and solve problems involving infinite series.

To find the formula for the nth partial sum of a series and determine its convergence, we start with the sequence defined by the terms \( a_n = \frac{1}{n(n+1)} \). This expression can be simplified using partial fraction decomposition. We can express \( a_n \) as:

\[ a_n = \frac{1}{n} - \frac{1}{n+1} \]

This decomposition allows us to observe a telescoping series when we write out the first few terms:

\[ a_1 = 1 - \frac{1}{2}, \quad a_2 = \frac{1}{2} - \frac{1}{3}, \quad a_3 = \frac{1}{3} - \frac{1}{4}, \quad a_4 = \frac{1}{4} - \frac{1}{5} \]

Next, we can calculate the nth partial sum \( S_n \) by summing these terms:

\[ S_n = a_1 + a_2 + a_3 + \ldots + a_n = \left( 1 - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \left( \frac{1}{3} - \frac{1}{4} \right) + \ldots + \left( \frac{1}{n} - \frac{1}{n+1} \right) \]

When we sum these terms, we notice that most terms cancel out:

\[ S_n = 1 - \frac{1}{n+1} \]

This shows that the nth partial sum can be expressed as:

\[ S_n = 1 - \frac{1}{n+1} \]

To determine if the series converges, we take the limit of \( S_n \) as \( n \) approaches infinity:

\[ \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( 1 - \frac{1}{n+1} \right) = 1 - 0 = 1 \]

Since the limit exists and is finite, we conclude that the series converges. Therefore, the series converges to the value of 1.

Understanding infinite series is crucial in mathematics, particularly when exploring special cases like geometric series. A geometric series is characterized by its form, which can be expressed as a summation of the first term \( a \) multiplied by the common ratio \( r \) raised to the power of \( n-1 \) or \( n \), depending on the starting index. The general form is:

$$ S = \sum_{n=1}^{\infty} a r^{n-1} $$

or

$$ S = \sum_{n=0}^{\infty} a r^{n} $$

In both cases, \( a \) represents the first term, and \( r \) is the common ratio. The convergence of a geometric series is determined by the absolute value of the ratio \( r \). If \( |r| < 1 \), the series converges; if \( |r| \geq 1 \), the series diverges.

When a geometric series converges, the sum can be calculated using the formula:

$$ S = \frac{a}{1 - r} $$

For example, consider a series where the first term \( a \) is 3 and the ratio \( r \) is \( \frac{2}{5} \). Since \( |r| = \frac{2}{5} < 1 \), the series converges. The sum can be calculated as follows:

$$ S = \frac{3}{1 - \frac{2}{5}} = \frac{3}{\frac{3}{5}} = 5 $$

In contrast, if we analyze a series where the terms are \( 1, \frac{3}{e}, \frac{3}{e^2}, \frac{3}{e^3}, \ldots \), we find that the common ratio \( r = \frac{3}{e} \). Given that \( e \approx 2.71 \), it follows that \( |r| > 1 \), indicating that this series diverges. This can be confirmed by observing that the ratio remains consistent when dividing consecutive terms, affirming the series' geometric nature.

Recognizing the characteristics of geometric series and applying the convergence tests are essential skills in calculus, enabling students to effectively analyze and compute sums of infinite series.

To determine the convergence of a series and find its sum, we can analyze the series using the properties of geometric series. A geometric series is defined as the sum from \( n = 0 \) to \( \infty \) of \( a \cdot r^n \), where \( a \) is the first term and \( r \) is the common ratio. The series converges if the absolute value of \( r \) is less than one, and diverges if \( |r| \geq 1 \).

In the first example, we consider the series:

\[ \sum_{n=0}^{\infty} \frac{3}{4^n} \]

We can factor out the constant \( 3 \) to rewrite the series as:

\[ 3 \sum_{n=0}^{\infty} \left(\frac{1}{4}\right)^n \]

Here, \( a = 3 \) and \( r = \frac{1}{4} \). Since \( |r| = \frac{1}{4} < 1 \), the series converges. The sum of a convergent geometric series is given by:

\[ S = \frac{a}{1 - r} \]

Substituting the values, we find:

\[ S = \frac{3}{1 - \frac{1}{4}} = \frac{3}{\frac{3}{4}} = 4 \]

Thus, the sum of the series is \( 4 \).

In the second example, we analyze the series:

\[ \sum_{n=0}^{\infty} \left(\frac{1}{2^n} + \frac{5}{3^n}\right) \]

This can be split into two separate series:

\[ \sum_{n=0}^{\infty} \frac{1}{2^n} + \sum_{n=0}^{\infty} \frac{5}{3^n} \]

For the first series, we have \( a = 1 \) and \( r = \frac{1}{2} \). Since \( |r| = \frac{1}{2} < 1 \), it converges:

\[ S_1 = \frac{1}{1 - \frac{1}{2}} = 2 \]

For the second series, \( a = 5 \) and \( r = \frac{1}{3} \). Again, since \( |r| = \frac{1}{3} < 1 \), it converges:

\[ S_2 = \frac{5}{1 - \frac{1}{3}} = \frac{5}{\frac{2}{3}} = \frac{15}{2} \]

Adding both sums gives:

\[ S = S_1 + S_2 = 2 + \frac{15}{2} = \frac{4}{2} + \frac{15}{2} = \frac{19}{2} \]

Thus, the total sum is \( \frac{19}{2} \).

In the third example, we are given a series explicitly:

\[ 9, -3, 1, -\frac{1}{3}, \ldots \]

To determine if this is a geometric series, we calculate the common ratio \( r \) by dividing consecutive terms:

\[ r = \frac{-3}{9} = -\frac{1}{3} \]

Checking the next terms confirms that \( r = -\frac{1}{3} \) consistently. Since \( |r| = \frac{1}{3} < 1 \), the series converges. The first term \( a = 9 \), so the sum is:

\[ S = \frac{a}{1 - r} = \frac{9}{1 - (-\frac{1}{3})} = \frac{9}{1 + \frac{1}{3}} = \frac{9}{\frac{4}{3}} = \frac{27}{4} \]

In summary, we have explored how to determine the convergence of geometric series and calculate their sums using the appropriate formulas. Understanding the conditions for convergence is crucial in analyzing series effectively.