## Table of Contents

- Infinite Series
- Convergence and Divergence Tests
- Comparison Test for Convergence
- How to Determine if a Series is Convergent or Divergent
- Convergence and Divergence Tests: Examples
- Lesson Summary

- Infinite Series
- Convergence and Divergence Tests
- Comparison Test for Convergence
- How to Determine if a Series is Convergent or Divergent
- Convergence and Divergence Tests: Examples
- Lesson Summary

An **infinite series** is the sum of an infinite sequence of numbers. Recall that a sequence is a list of numbers, often appearing as {eq}\{ a_n \} = a_1,a_2,a_3,...,a_n {/eq}. A series a sum of a sequence of terms, often appearing as {eq}\sum a_n = a_1+ a_2+ a_3+...+a_n {/eq}.

Since it can be difficult to model behavior of an infinite sum of numbers, it is helpful to consider **partial sums**. Partial sums are finite series, denoted {eq}S_n = a_1+a_2+a_3+...+a_n = \sum_{k=1}^n a_k {/eq}. The expression {eq}S_n {/eq} is called the n-th partial sum.

An infinite series can be found by taking the limit of the sequence of partial sums. Then, the infinite series is equal to {eq}\lim_{n \rightarrow \infty} S_n = \lim_{n \rightarrow \infty} \sum_{k=1}^n a_k {/eq}. The last part of the equation, {eq}\lim_{n \rightarrow \infty} \sum_{k=1}^n a_k {/eq}, is called the **sigma notation** for an infinite series. Sometimes notation is simplified, and infinite series are written as {eq}\sum_{k=1}^\infty a_k {/eq}.

Some examples of infinite series are as follows.

Let {eq}S_n = \frac{2^n -1}{2^n} {/eq}. Find the infinite series {eq}\lim_{n \rightarrow \infty} S_n {/eq}

- Process

Notice that {eq}S_n = \frac{2^n - 1}{2^n} = 1 - \frac{1}{2^n} {/eq}. Next, calculate some of the partial sums, looking for the behavior of {eq}S_n {/eq} as n increases without bound.

{eq}S_1 = 1 - \frac{1}{2} = \frac{1}{2} \\ S_2 = 1 - \frac{1}{4} = \frac{3}{4}\\ S_3 = 1 - \frac{1}{8} = \frac{7}{8}\\ S_4 = 1 - \frac{1}{16} = \frac{15}{16}\\ {/eq}

Notice that as n increases, {eq}S_n {/eq} tends toward 1. Then, the limit of the sequence of partial sums is 1.

- Solution

{eq}\lim_{n \rightarrow \infty} S_n = 1 {/eq}

Let {eq}S_n = n {/eq}. Find the infinite series {eq}\lim_{n \rightarrow \infty} S_n {/eq}

- Process

Notice that {eq}S_1 = 1, S_2 = 2, S_3 = 3 {/eq}.

As n increases, {eq}S_n {/eq} also increases without bounds. Then, the sequence of partial sums tends toward infinity.

- Solution

{eq}\lim_{n \rightarrow \infty} \sum_{k=1}^n k = \infty {/eq}

An infinite series will either converge to a real number, diverge to positive or negative infinity, or oscillate. The series' behavior can be found by taking the limit of the sequence of partial sums. If the limit of the sequence of partial sums is equal to a real number, the series is called a **convergent series** and is denoted {eq}\sum_{n=1}^\infty a_n = L {/eq}, where L is the real-valued sum. If the sequence of partial sums does not converge, the series is called a **divergent series**. Divergent series include infinite series that diverge to positive infinity, diverge to negative infinity, and oscillate.

**Geometric series** are convergent series that take the form {eq}\sum_{k=1}^\infty ar^{k-1} = a+ar+ar^2+a^3+... {/eq}, where {eq}a \neq 0 {/eq} and {eq}r \neq 1 {/eq}. Infinite series that can be expressed as geometric series converge to {eq}\frac{a}{1-r} {/eq}.

**P-series **are infinite series in the form {eq}\sum_{n=1}^\infty \frac{1}{n^p} {/eq}. The rule for p-series is that the infinite series diverges if {eq}p \leq 1 {/eq} and converges if {eq}p > 1 {/eq}. A series with {eq}p=1 {/eq} is called a harmonic series, which is a well-known divergent series that can be written as {eq}\sum_{n =1}^\infty \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... {/eq}.

Viewing the graph of the harmonic series, the sequence clearly converges to zero. The sum of the sequence, however, can be viewed in the graph as the area beneath the function. This area, while growing slowly, diverges to infinity.

A **telescoping sum** is an infinite series with middle terms that cancel, leaving only the first and last terms in the partial sums. For example, consider the infinite series {eq}\sum_{n =1}^\infty (\frac{1}{n} - \frac{1}{n+1}) {/eq}. This series expands to:

{eq}\sum_{n =1}^\infty (\frac{1}{n} - \frac{1}{n+1}) = (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) +(\frac{1}{3} - \frac{1}{4}) + ... \\ = \lim_{n \rightarrow \infty} [ (1 - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) +(\frac{1}{3} - \frac{1}{4}) + ... +(\frac{1}{n-1} - \frac{1}{n}) + (\frac{1}{n} - \frac{1}{n+1}) ] {/eq}

Notice that the middle numbers repeat negative and then positive, cancelling each set of values to zero.

{eq}\sum_{n =1}^\infty (\frac{1}{n} - \frac{1}{n+1}) = \lim_{n \rightarrow \infty} (1 - \frac{1}{n+1}) = 1 {/eq}

The last step comes from incorporating the limit at infinity, which cancels {eq}\frac{1}{n+1} {/eq} to zero. Then, this example of a telescoping sum converges to 1.

Oscillating series are infinite series with partial sums that oscillate between two or more values. Consider the infinite series {eq}\sum_{n=1}^\infty (-1)^{n-1} = 1-1+1-1+1-1+... {/eq}. Notice that:

{eq}S_1 = 1\\ S_2 = 1-1 = 0\\ S_3 = 1-1+1 = 1\\ S_4 = 1-1+1-1 =0 {/eq}

Since the sequence of the partial sums oscillates, the limit of the sequence of partial sums does not exist. Then, the infinite series is not convergent. Since an infinite series that is not convergent is divergent, oscillating series are divergent series.

Oscillating series are less common than other forms and are not used very frequently in testing methodologies.

Many series do not fit the exact form of geometric series, oscillating series, p-series, or telescoping sums; one way to discern the behavior of series is to use **convergence and divergence tests**. The difference between the two types of tests is that divergence tests provide certain conditions for divergent series, while convergence tests provide certain conditions for convergent series. Divergence tests can never test for convergence, and convergence tests can never test for divergence.

The divergence test is sometimes called the **series divergence test** or the **basic divergence test**. This test states that {eq}\sum_{n=1}^\infty a_n {/eq} diverges if {eq}\lim_{n\rightarrow \infty} a_n \neq 0 {/eq} or if the limit does not exist.

Note that if {eq}\lim_{n\rightarrow \infty} a_n = 0 {/eq}, the divergence test is inconclusive. In these cases, the series could be convergent, or it could still be divergent.

The **absolute convergence test** states that if the sum of a sequence, {eq}\sum_{n=1}^\infty | a_n | {/eq} converges, then the sum of the sequence {eq}\sum_{n=1}^\infty a_n {/eq} also converges.

Note that if {eq}\sum_{n=1}^\infty | a_n | {/eq} does not converge, the test is inconclusive and does not give any information about {eq}\sum_{n=1}^\infty a_n {/eq}.

If the convergence and divergence tests fail, the next step is to apply a **comparison test for convergence**. Comparison tests use well-known series and compare them to unknown series, allowing the evaluation of infinite series that do not fit common forms such as geometric series and p-series.

Let {eq}\sum_{n=1}^\infty a_n {/eq} be the series in question. Let {eq}\sum_{n=1}^\infty a_n {/eq} be a series with a known solution. Let every element of each series be positive - {eq}a_n, b_n >0 {/eq}. Then, the **comparison test** states:

- If {eq}\sum_{n=1}^\infty b_n {/eq} converges and {eq}a_n \leq b_n {/eq} for all n, then {eq}\sum_{n=1}^\infty a_n {/eq} converges.

- If {eq}\sum_{n=1}^\infty b_n {/eq} diverges and {eq}a_n \geq b_n {/eq} for all n, then {eq}\sum_{n=1}^\infty a_n {/eq} diverges.

Note that to use the comparison test, there must be two series with all positive terms, and one series must be smaller than the other. The comparison test works by using the fact that a series will converge if it is smaller than another series that converges, and a series will diverge if it is larger than another series that diverges. Then to use this test to determine convergence of a series, it is necessary to have another larger convergent series; to use this test to determine divergence of a series, it is necessary to have another smaller divergent series.

Let {eq}\sum_{n=1}^\infty a_n {/eq} be the series in question. Let {eq}\sum_{n=1}^\infty a_n {/eq} be a series with a known solution. Let every element of each series be positive - {eq}a_n, b_n >0 {/eq}. Then, the **limit comparison test** states:

- If {eq}\lim_{n \rightarrow \infty} \frac{a_n}{b_n} > 0 {/eq}, then {eq}\sum_{n=1}^\infty a_n {/eq} has the same convergent or divergent behavior as {eq}\sum_{n=1}^\infty b_n {/eq}.

- If {eq}\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = 0 {/eq} and {eq}\sum_{n=1}^\infty b_n {/eq} converges, then {eq}\sum_{n=1}^\infty a_n {/eq} also converges.

- If {eq}\lim_{n \rightarrow \infty} \frac{a_n}{b_n} = \infty {/eq} and {eq}\sum_{n=1}^\infty b_n {/eq} diverges, then {eq}\sum_{n=1}^\infty a_n {/eq} also diverges.

With so many testing options, it can be difficult to know how to determine if a series is convergent or divergent. Since series have clear forms, check the given series to see if it has the form of a geometric series ({eq}\sum ar^n {/eq}), p-series ({eq}\sum \frac{1}{n^p} {/eq}), or telescoping sum (middle terms cancel, typically involves subtraction). If the given infinite series does not fit into any known forms, series tests are needed.

Start with one of the easiest tests to apply: the divergence test. If the limit of the sequence within the sum does not equal zero, the series must be divergent. If the limit of the sequence within the sum does equal zero, the divergence test is inconclusive and other tests must be applied.

The absolute convergence test is helpful when the terms within the series oscillate between positive or negative values. The absolute convergence test converts all of the values positive and states that if the resulting series converges, then the version with negative values must also converge.

The comparison test works when there is a known form that the given series is very close to. Typically, the original series will be compared to a geometric series or p-series.

The limit comparison test is a more general form of the comparison test and works for a wider variety of series but is slightly more difficult to apply. Attempt the limit comparison test last if the other tests have failed.

When a problem gives an infinite series with no special instructions, cycle through the known forms and tests until the series is found to be divergent or convergent. Some examples are as follows.

Determine whether the infinite series {eq}\sum_{k=0}^\infty (\frac{3}{5})^k {/eq} converges or diverges.

- Process

Notice that this series has the form of a geometric series:

{eq}\sum_{k=0}^\infty (\frac{3}{5})^k = 1 + \frac{3}{5} + (\frac{3}{5})^2 + (\frac{3}{5})^2 + ... {/eq}

In this case, {eq}a = 1 {/eq} and {eq}r = \frac{3}{5} {/eq}. Then, the series converges to {eq}\frac{a}{1-r} = \frac{1}{1-\frac{3}{5}} = \frac{1}{\frac{2}{5}} = \frac{5}{2} {/eq}.

- Solution

{eq}\sum_{k=0}^\infty (\frac{3}{5})^k {/eq} converges to {eq}\frac{5}{2} {/eq}.

Determine whether the infinite series {eq}\sum_{n=1}^\infty (\frac{n+1}{n}) {/eq} converges or diverges.

- Process

Notice that {eq}\sum_{n=1}^\infty (\frac{n+1}{n}) = \sum_{n=1}^\infty (1 + \frac{1}{n}) {/eq}. Then, the given series is very similar to the harmonic series (p-series with p = 1), and the comparison test can be applied. Since {eq}1+ \frac{1}{n} > \frac{1}{n} {/eq} for all n and {eq}\sum_{n=1}^\infty \frac{1}{n} {/eq} diverges, then {eq}\sum_{n=1}^\infty (1+ \frac{1}{n}) {/eq} also diverges.

- Solution

{eq}\sum_{n=1}^\infty (\frac{n+1}{n}) {/eq} diverges.

Determine whether the infinite series {eq}\sum_{n=1}^\infty (\frac{n^3}{n^5}) {/eq} converges or diverges.

- Process

Notice that {eq}\sum_{n=1}^\infty (\frac{n^3}{n^5}) = \sum_{n=1}^\infty (\frac{1}{n^2}) {/eq}. The resulting series is a p-series with p > 1. Then, the series converges.

- Solution

{eq}\sum_{n=1}^\infty (\frac{n^3}{n^5}) {/eq} converges.

Determine whether the infinite series {eq}\sum_{k=0}^\infty (\frac{3n^2 + 7n +1}{4n^2-6n+2}) {/eq} converges or diverges.

- Process

Notice that this infinite series does not fit the form of a geometric series, p-series, or telescoping sum. Then, the divergence test is a quick and easy test that can be applied next. Consider:

{eq}\lim_{n \rightarrow \infty} (\frac{3n^2 + 7n +1}{4n^2-6n+2}) = \frac{3}{4} \neq 0 {/eq}

Then, the series diverges by the divergence test.

NOTE: The limit was simplified using the rules of limits at infinity (3/4 is the horizontal asymptote of the function within the limit).

- Solution

{eq}\sum_{k=0}^\infty (\frac{3n^2 + 7n +1}{4n^2-6n+2}) {/eq} diverges.

**Infinite series **are limits of sequences of **partial sums**, denoted {eq}\lim_{n \rightarrow \infty} S_n {/eq}. In general, a series is a sum of a sequence of terms. In **sigma notation**, infinite series are written as {eq}\lim_{n \rightarrow \infty} \sum_{k=1}^n a_k {/eq} or {eq}\sum_{n=1}^\infty a_n {/eq}. An infinite series will either be convergent or divergent. A **convergent series** converges to a real number and is denoted {eq}\sum_{n=1}^\infty a_n = L {/eq}, where L is the real-valued sum. A **divergent series** is a series that is not convergent.

The process of how to determine if a series is convergent or divergent involves analyzing the form of infinite series, applying **convergence and divergence tests**, and utilizing **comparison tests for convergence**.

Known forms of infinite series:

**Geometric series****P-series****Telescoping sum**

Tests for convergence and divergence:

**Basic/series divergence test****Absolute convergence test**

Comparison tests:

**Comparison test****Limit comparison test**

To use the comparison test, two series must have terms that are all positive and one series must be smaller than the other. When using the comparison test for convergence, there must be a larger convergent series; when using the comparison test for divergence, there must be a smaller divergent series.

When a given infinite series is unknown, cycle through the convergence, divergence, and comparison tests to determine whether the series is convergent or divergent.

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Frequently Asked Questions

Divergence testing only tests for divergence, while convergence testing only tests for convergence.

A divergence test will never reveal if a series is convergent, and a convergence test will never reveal if a series is divergent.

First, see if the given series fits any common forms of series. The series may be a geometric series, a p-series, or a telescoping sum.

It the series does not fit a known form, begin applying divergence, convergence, and comparison tests. Begin with the divergence test. Then, check if the absolute convergence test might be applicable. If the series is similar to a known form, use the comparison test. If no other test has worked, try the limit comparison test.

Some tests will determine if a series is convergent or divergent- these tests include the limit comparison test and the comparison test. The comparison tests require the given series to be compared to a series with known convergent or divergent behavior.

Other tests will only determine convergence or only determine divergence. The divergence test provides a condition for divergent series, but the test fails if the given series does not meet this condition. The absolute convergence test provides a condition for convergent series, but the test fails if the given series does not meet this condition.

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