The Central Idea
For a series with nonnegative terms,
\[ \sum_{n=1}^{\infty} a_n, \qquad a_n\ge 0, \]the partial sums can only stay the same or increase. Therefore, a positive-term series converges exactly when its partial sums stay bounded above.
Convergence tests give us indirect ways to establish that boundedness—or to prove that the partial sums must grow without bound.
Why Positive Terms Are Special
If every term satisfies \(a_n\ge0\), then
\[ S_{N+1}=S_N+a_{N+1}\ge S_N. \]Thus the sequence of partial sums \((S_N)\) is monotone increasing. A basic convergence theorem for sequences then tells us:
\[ \boxed{\text{positive-term series converges} \iff (S_N)\text{ is bounded above}.} \]Positive-Term and Eventually Positive Series
A series does not need to be positive from its very first term. If there is some index \(N_0\) such that
\[ a_n\ge0 \qquad \text{for all } n\ge N_0, \]then the tail is positive-term. Since changing or removing finitely many terms does not change convergence, the positive-term tests can be applied to that tail.
First Screening Test: Do the Terms Go to Zero?
Before using a more specialized test, always check
\[ \lim_{n\to\infty}a_n. \]If this limit is not zero, then
\[ \boxed{\sum a_n \text{ diverges}.} \]This is the divergence test, also called the nth-term test for divergence.
Zero Terms Are Necessary but Not Sufficient
The condition \(a_n\to0\) is necessary for convergence, but it does not prove convergence. The harmonic series
\[ \sum_{n=1}^{\infty}\frac1n \]has terms approaching zero, yet the series diverges.
Benchmark Series
Most comparison tests rely on a small collection of familiar series whose behavior is already known. The two most important starting benchmarks are geometric series and \(p\)-series.
Geometric Benchmark
A geometric series
\[ \sum_{n=0}^{\infty} ar^n \]converges exactly when
\[ \boxed{|r|<1}. \]For positive-term comparison work, a common benchmark is
\[ \sum_{n=0}^{\infty} r^n,\qquad 0The \(p\)-Series Benchmark
A \(p\)-series has the form
\[ \sum_{n=1}^{\infty}\frac1{n^p}. \]Its convergence behavior is
\[ \boxed{\sum_{n=1}^{\infty}\frac1{n^p} \begin{cases} \text{converges}, & p>1,\\ \text{diverges}, & p\le1. \end{cases}} \]The Harmonic Series
The boundary case \(p=1\) is the harmonic series:
\[ \sum_{n=1}^{\infty}\frac1n. \]It diverges. This fact makes \(1/n\) one of the most useful divergent comparison benchmarks.
Why \(p=1\) Is a Threshold
Terms like \(1/n^2\) shrink fast enough for the total accumulated amount to remain finite, while \(1/n\) shrinks too slowly. The integral test will make this threshold precise.
The Integral Test
Suppose \(f\) is positive, continuous, and decreasing for all sufficiently large \(x\), and
\[ a_n=f(n). \]Then
\[ \boxed{\sum_{n=N}^{\infty}a_n \text{ and } \int_N^{\infty}f(x)\,dx \text{ have the same convergence behavior}.} \]What the Integral Test Does Not Say
The integral test compares convergence behavior. It does not generally say that
\[ \sum_{n=N}^{\infty}f(n)=\int_N^{\infty}f(x)\,dx. \]Those are different quantities. Their convergence or divergence agrees under the test's hypotheses.
Why Decreasing Matters
When \(f\) is positive and decreasing, rectangles built from the values \(f(n)\) can be placed consistently above or below the graph. This produces inequalities linking a series tail with an improper integral.
Integral Test and \(p\)-Series
Take
\[ f(x)=\frac1{x^p}. \]For \(p\ne1\),
\[ \int_1^{\infty}x^{-p}\,dx \]converges exactly when \(p>1\). For \(p=1\),
\[ \int_1^{\infty}\frac{dx}{x} \]diverges. Therefore the integral test produces the \(p\)-series criterion.
Example: Integral Test
Determine whether
\[ \sum_{n=1}^{\infty}\frac1{(n+2)^2} \]converges. Let
\[ f(x)=\frac1{(x+2)^2}. \]The function is positive, continuous, and decreasing for \(x\ge1\), and
\[ \int_1^{\infty}\frac{dx}{(x+2)^2}=\frac13. \]The improper integral converges, so the series converges.
A Logarithmic Integral Benchmark
A useful family is
\[ \sum_{n=2}^{\infty}\frac1{n(\ln n)^p}. \]The substitution \(u=\ln x\) in the corresponding improper integral gives
\[ \int\frac{dx}{x(\ln x)^p}=\int u^{-p}\,du. \]Therefore this series converges for \(p>1\) and diverges for \(p\le1\).
Direct Comparison Test
Suppose
\[ 0\le a_n\le b_n \]for all sufficiently large \(n\). Then two one-way conclusions are especially useful.
Comparison for Convergence
If
\[ 0\le a_n\le b_n \]and
\[ \sum b_n \text{ converges}, \]then
\[ \boxed{\sum a_n \text{ converges}.} \]A smaller positive series cannot accumulate more than a larger convergent one.
Comparison for Divergence
If
\[ 0\le b_n\le a_n \]and
\[ \sum b_n \text{ diverges}, \]then
\[ \boxed{\sum a_n \text{ diverges}.} \]A larger positive series cannot remain finite when a smaller one already grows without bound.
The Direction of Comparison Matters
To prove convergence, compare your series above by a known convergent series. To prove divergence, compare your series below by a known divergent series.
Example: Direct Comparison for Convergence
Consider
\[ \sum_{n=1}^{\infty}\frac1{n^2+3n}. \]Since \(n^2+3n\ge n^2\),
\[ 0<\frac1{n^2+3n}\le\frac1{n^2}. \]Because \(\sum 1/n^2\) converges, the given series converges.
Example: Direct Comparison for Divergence
Consider
\[ \sum_{n=1}^{\infty}\frac{n}{n^2+1}. \]For \(n\ge1\), \(n^2+1\le2n^2\), so
\[ \frac{n}{n^2+1}\ge\frac1{2n}. \]Since \(\sum 1/(2n)\) diverges, the given series diverges.
How to Invent a Comparison
For rational expressions involving powers of \(n\), focus first on the highest-degree terms. For example,
\[ \frac{3n+7}{n^3+4} \]behaves roughly like
\[ \frac{3n}{n^3}=\frac3{n^2}. \]This suggests comparing with a convergent \(p\)-series.
Dominant Powers
For large \(n\), lower-degree terms usually matter much less than the highest powers. A rational term
\[ \frac{P(n)}{Q(n)} \]with polynomial numerator and denominator often behaves like a constant multiple of
\[ n^{\deg P-\deg Q}. \]Limit Comparison Test
Let \(a_n>0\) and \(b_n>0\) for all sufficiently large \(n\). If
\[ \lim_{n\to\infty}\frac{a_n}{b_n}=L, \qquad 0Why Limit Comparison Works
If \(a_n/b_n\to L\) with \(L>0\), then for large \(n\), \(a_n\) is trapped between positive constant multiples of \(b_n\). The two series therefore have the same large-scale accumulation behavior.
Example: Limit Comparison with \(1/n^2\)
Consider
\[ \sum_{n=1}^{\infty}\frac{5n+2}{n^3+7}. \]Compare with \(b_n=1/n^2\):
\[ \frac{a_n}{b_n} =\frac{(5n+2)n^2}{n^3+7} \longrightarrow 5. \]Since \(\sum 1/n^2\) converges, the given series converges.
Example: Limit Comparison with the Harmonic Series
Consider
\[ \sum_{n=1}^{\infty}\frac{4n^2+1}{n^3+5}. \]Compare with \(b_n=1/n\):
\[ \frac{a_n}{b_n} =\frac{n(4n^2+1)}{n^3+5} \longrightarrow4. \]Since the harmonic series diverges, the given series diverges.
Direct Comparison or Limit Comparison?
Use direct comparison when a clean inequality is easy to prove. Use limit comparison when the terms clearly have the same dominant behavior but an exact inequality would be awkward.
One-Way Limit Comparisons
Some useful conclusions remain when the limit is \(0\) or \(\infty\).
- If \(a_n/b_n\to0\) and \(\sum b_n\) converges, then \(\sum a_n\) converges.
- If \(a_n/b_n\to\infty\) and \(\sum b_n\) diverges, then \(\sum a_n\) diverges.
These are one-way statements, not full equivalences.
The Ratio Test
For positive terms, define
\[ L=\lim_{n\to\infty}\frac{a_{n+1}}{a_n}, \]when the limit exists. Then
\[ \boxed{ \begin{aligned} L<1 &\Rightarrow \sum a_n \text{ converges},\\ L>1 &\Rightarrow \sum a_n \text{ diverges},\\ L=1 &\Rightarrow \text{the test is inconclusive}. \end{aligned}} \]Why the Ratio Test Resembles a Geometric Series
If \(a_{n+1}/a_n\) approaches a number smaller than \(1\), then eventually the terms shrink by approximately a fixed multiplicative factor. That is the same kind of behavior that makes a geometric series converge.
Ratio Test: Factorials
The ratio test is especially effective when factorials appear, because
\[ (n+1)!=(n+1)n!. \]Consecutive-term division usually causes most of the factorial expression to cancel.
Example: Ratio Test with a Factorial
Consider
\[ \sum_{n=0}^{\infty}\frac{3^n}{n!}. \]Then
\[ \frac{a_{n+1}}{a_n} =\frac{3^{n+1}}{(n+1)!}\frac{n!}{3^n} =\frac3{n+1}\longrightarrow0. \]Since \(L=0<1\), the series converges.
Ratio Test: Exponential Powers
Expressions containing \(c^n\), factorials, or products indexed by \(n\) often simplify dramatically under \(a_{n+1}/a_n\). This makes the ratio test a natural first choice for many such series.
Example: Ratio Test with Polynomial and Exponential Terms
Consider
\[ \sum_{n=1}^{\infty}\frac{n^3}{4^n}. \]The ratio is
\[ \frac{a_{n+1}}{a_n} =\frac{(n+1)^3}{4^{n+1}}\frac{4^n}{n^3} =\frac14\left(1+\frac1n\right)^3 \longrightarrow\frac14. \]Therefore the series converges.
When the Ratio Test Gives \(L>1\)
If \(L>1\), the terms eventually fail to shrink toward zero fast enough—in fact they eventually grow in a ratio sense. The series therefore diverges.
Ratio Test Inconclusive at \(L=1\)
Both
\[ \sum\frac1n \qquad\text{and}\qquad \sum\frac1{n^2} \]produce ratio-test limit \(L=1\), yet the first diverges and the second converges. Therefore \(L=1\) gives no conclusion.
The Root Test
For positive terms, define
\[ L=\lim_{n\to\infty}\sqrt[n]{a_n}, \]when the limit exists. Then
\[ \boxed{ \begin{aligned} L<1 &\Rightarrow \sum a_n \text{ converges},\\ L>1 &\Rightarrow \sum a_n \text{ diverges},\\ L=1 &\Rightarrow \text{the test is inconclusive}. \end{aligned}} \]Why the Root Test Works
The quantity \(\sqrt[n]{a_n}\) measures the effective exponential base of the term. If that base approaches a number below \(1\), the terms behave roughly like a convergent geometric sequence.
Root Test: Terms Raised to the \(n\)th Power
The root test is especially useful when the general term contains a large expression raised to the power \(n\), because the \(n\)th root removes that outer exponent immediately.
Example: Root Test
Consider
\[ \sum_{n=1}^{\infty}\left(\frac{2n+1}{5n+4}\right)^n. \]Then
\[ \sqrt[n]{a_n}=\frac{2n+1}{5n+4}\longrightarrow\frac25. \]Since \(2/5<1\), the series converges.
Root Test Inconclusive at \(L=1\)
As with the ratio test, \(L=1\) gives no decision. For example,
\[ \sqrt[n]{\frac1{n^2}}=n^{-2/n}\longrightarrow1, \]even though \(\sum1/n^2\) converges.
Ratio Test versus Root Test
The ratio test is usually preferable for factorials and products involving consecutive indices. The root test is often preferable for expressions of the form
\[ [g(n)]^n. \]Either test may work in some problems; choose the one that simplifies the algebra most directly.
Test Selection Is Pattern Recognition
There is no single best convergence test for every series. Efficient work begins by recognizing the visible structure before calculating anything.
A Practical Test-Selection Guide
- Recognizable geometric form: use the geometric-series criterion.
- Recognizable telescoping form: inspect partial sums directly.
- Rational powers of \(n\): compare with a \(p\)-series.
- A term naturally extends to a positive decreasing function: consider the integral test.
- Dominant behavior is obvious but inequalities are messy: use limit comparison.
- Factorials or strong exponential factors: try the ratio test.
- An entire expression is raised to the \(n\)th power: try the root test.
Start with the Simplest Test
A complicated theorem is not automatically a better theorem. If direct comparison settles the problem in two lines, there is little value in forcing a ratio or integral calculation.
Finite Changes Do Not Affect Convergence
Convergence depends on the tail. Therefore a comparison or monotonicity condition only needs to hold for all sufficiently large \(n\). A small number of exceptional early terms does not change the final convergence result.
Constants Usually Do Not Affect Convergence
If \(c>0\), then
\[ \sum a_n \quad\text{and}\quad \sum c a_n \]have the same convergence behavior. Constant factors matter for the value of a sum, but not for whether a positive-term series is finite or infinite.
Example: Ignoring Harmless Constants
The series
\[ \sum_{n=1}^{\infty}\frac{17}{n^3} \]converges because \(\sum1/n^3\) is a convergent \(p\)-series. The factor \(17\) does not change convergence.
Asymptotic Thinking
Many convergence decisions reduce to identifying what a term behaves like for large \(n\). Informally,
\[ a_n\sim b_n \]means \(a_n/b_n\to1\). Positive series with asymptotically equivalent terms share convergence behavior by limit comparison.
Example: Asymptotic Equivalence
For
\[ a_n=\frac{n+4}{n^3+2n}, \]the dominant behavior is
\[ a_n\sim\frac{n}{n^3}=\frac1{n^2}. \]This strongly suggests limit comparison with \(1/n^2\), which confirms convergence.
Polynomial-over-Polynomial Shortcut
For a positive rational term whose leading behavior is
\[ \frac{Cn^m}{Dn^k}, \]think first of
\[ \frac{C}{D}\frac1{n^{k-m}}. \]The exponent \(k-m\) points toward the relevant \(p\)-series benchmark.
Example: Degree Difference Greater Than One
Consider
\[ \sum_{n=1}^{\infty}\frac{2n^2+1}{n^5+3n}. \]The term behaves like
\[ \frac{2n^2}{n^5}=\frac2{n^3}. \]Since \(p=3>1\), limit comparison with \(1/n^3\) shows convergence.
Example: Degree Difference Equal to One
Consider
\[ \sum_{n=1}^{\infty}\frac{3n^2+7}{n^3+1}. \]The term behaves like
\[ \frac3n. \]Limit comparison with the harmonic series shows divergence.
Example: Terms Do Not Approach Zero
Consider
\[ \sum_{n=1}^{\infty}\frac{n^2+1}{2n^2+3}. \]The terms satisfy
\[ \frac{n^2+1}{2n^2+3}\longrightarrow\frac12\ne0. \]The divergence test settles the problem immediately. No further test is needed.
Common Error: Using the Divergence Test Backward
From \(a_n\to0\), you may not conclude that \(\sum a_n\) converges. The test only gives a guaranteed conclusion when the term limit is nonzero or fails to exist.
Common Error: Comparing in the Wrong Direction
If \(0\le a_n\le b_n\) and \(\sum b_n\) diverges, nothing follows. A smaller series may converge or diverge. Likewise, if \(b_n\le a_n\) and \(\sum b_n\) converges, nothing follows about the larger series.
Common Error: Comparing with an Unknown Benchmark
A comparison only helps when the comparison series has known convergence behavior. Replacing one difficult series with another equally unfamiliar series does not solve the problem.
Common Error: Forgetting Integral-Test Conditions
Before invoking the integral test, verify that the corresponding function is positive, continuous, and decreasing on a sufficiently large interval. These hypotheses are part of the theorem.
Common Error: Confusing the Integral with the Sum
The integral test determines shared convergence behavior. It does not usually evaluate the series itself.
Common Error: Treating \(L=1\) as Divergence
For both the ratio and root tests,
\[ L=1 \]means only that the test is inconclusive. You must choose another method.
Common Error: Doing Too Much Algebra
Convergence questions often need only large-\(n\) behavior. Exact simplification is useful when it reveals structure, but expanding every term or finding a closed form is often unnecessary.
A Reliable Positive-Series Workflow
- Check whether \(a_n\to0\). If not, stop: the series diverges.
- Look for exact geometric or telescoping structure.
- Identify the dominant large-\(n\) behavior.
- For powers of \(n\), compare with a \(p\)-series.
- Use direct comparison when a clean inequality is available.
- Use limit comparison when terms have matching asymptotic form.
- Use the integral test when a natural positive decreasing function is present.
- Use the ratio test for factorial or strong exponential structure.
- Use the root test for expressions raised to the \(n\)th power.
- If a test is inconclusive, switch tests rather than forcing it.
Example: Choosing Between Tests
For
\[ \sum_{n=1}^{\infty}\frac{n^4}{7^n}, \]comparison is possible, but the ratio test is more direct because the exponential factor simplifies immediately. The ratio limit is \(1/7\), so the series converges.
Example: Choosing Limit Comparison
For
\[ \sum_{n=1}^{\infty}\frac{7n^3-2n}{4n^5+9}, \]the dominant behavior is \((7/4)n^{-2}\). Limit comparison with \(1/n^2\) is shorter than constructing a detailed direct inequality.
Example: Choosing the Root Test
For
\[ \sum_{n=1}^{\infty}\left(\frac{3n+1}{4n+7}\right)^n, \]the \(n\)th root removes the outer exponent at once:
\[ \sqrt[n]{a_n}=\frac{3n+1}{4n+7}\to\frac34<1. \]The series converges.
Example: One Test Fails, Another Works
For the \(p\)-series
\[ \sum_{n=1}^{\infty}\frac1{n^3}, \]the ratio test gives \(L=1\), so it is inconclusive. Recognizing the series as a \(p\)-series immediately proves convergence because \(p=3>1\).
Convergence Is Not the Same as Finding the Sum
Most convergence tests answer only the yes-or-no question: does the infinite series converge? They usually do not produce an exact value for the sum.
Why This Distinction Matters
A geometric series may yield both convergence and an exact sum. A comparison test may prove convergence without providing the sum at all. In analysis, knowing that a finite limit exists is often valuable even when no simple closed form is available.
Practice 1: \(p\)-Series Recognition
Determine whether the series converges:
\[ \sum_{n=1}^{\infty}\frac1{n^{5/4}}. \]Practice 2: Divergence Test
Determine whether the series converges:
\[ \sum_{n=1}^{\infty}\frac{3n+1}{5n+2}. \]Practice 3: Direct Comparison
Determine whether the series converges:
\[ \sum_{n=1}^{\infty}\frac1{n^2+9}. \]Practice 4: Divergent Comparison
Determine whether the series converges:
\[ \sum_{n=1}^{\infty}\frac{n}{n^2+4}. \]Practice 5: Limit Comparison
Determine whether the series converges:
\[ \sum_{n=1}^{\infty}\frac{6n^2+1}{n^4+3}. \]Practice 6: Integral Test
Determine whether the series converges:
\[ \sum_{n=2}^{\infty}\frac1{n(\ln n)^2}. \]Practice 7: Ratio Test
Determine whether the series converges:
\[ \sum_{n=1}^{\infty}\frac{5^n}{n!}. \]Practice 8: Ratio-Test Inconclusive
Apply the ratio test to
\[ \sum_{n=1}^{\infty}\frac1{n^4} \]and then determine convergence by a better method.
Practice 9: Root Test
Determine whether the series converges:
\[ \sum_{n=1}^{\infty}\left(\frac{n+2}{3n+1}\right)^n. \]Practice 10: Choose the Best Test
Choose an efficient convergence test and determine whether
\[ \sum_{n=1}^{\infty}\frac{4n^3+7}{n^4+2n} \]converges or diverges.
Practice Answers
Practice 1
This is a \(p\)-series with \(p=5/4>1\). Therefore,
\[ \boxed{\text{the series converges}.} \]Practice 2
The terms satisfy
\[ \frac{3n+1}{5n+2}\to\frac35\ne0. \]Therefore,
\[ \boxed{\text{the series diverges}.} \]Practice 3
Since \(n^2+9\ge n^2\),
\[ 0<\frac1{n^2+9}\le\frac1{n^2}. \]The comparison series converges, so
\[ \boxed{\text{the given series converges}.} \]Practice 4
For \(n\ge2\), \(n^2+4\le2n^2\), so
\[ \frac{n}{n^2+4}\ge\frac1{2n}. \]Since \(\sum1/(2n)\) diverges,
\[ \boxed{\text{the given series diverges}.} \]Practice 5
Compare with \(b_n=1/n^2\):
\[ \frac{a_n}{b_n} =\frac{n^2(6n^2+1)}{n^4+3}\to6. \]Since \(\sum1/n^2\) converges,
\[ \boxed{\text{the given series converges}.} \]Practice 6
Use
\[ f(x)=\frac1{x(\ln x)^2}. \]It is positive, continuous, and decreasing for sufficiently large \(x\). With \(u=\ln x\),
\[ \int_2^{\infty}\frac{dx}{x(\ln x)^2} =\int_{\ln2}^{\infty}u^{-2}\,du, \]which converges. Hence
\[ \boxed{\text{the series converges}.} \]Practice 7
With \(a_n=5^n/n!\),
\[ \frac{a_{n+1}}{a_n}=\frac5{n+1}\to0<1. \]Therefore,
\[ \boxed{\text{the series converges}.} \]Practice 8
The ratio test gives
\[ \frac{a_{n+1}}{a_n}=\left(\frac{n}{n+1}\right)^4\to1, \]so the ratio test is inconclusive. But the series is a \(p\)-series with \(p=4>1\), so
\[ \boxed{\text{the series converges}.} \]Practice 9
The root test gives
\[ \sqrt[n]{a_n}=\frac{n+2}{3n+1}\to\frac13<1. \]Therefore,
\[ \boxed{\text{the series converges}.} \]Practice 10
The dominant behavior is
\[ \frac{4n^3}{n^4}=\frac4n. \]Compare with \(b_n=1/n\):
\[ \frac{a_n}{b_n} =\frac{n(4n^3+7)}{n^4+2n}\to4. \]Since the harmonic series diverges,
\[ \boxed{\text{the given series diverges}.} \]What to Remember
Positive-term series are governed by monotone partial sums. The main task is to decide whether those partial sums remain bounded.
- Always screen with the term limit first.
- Know the geometric and \(p\)-series benchmarks.
- Use direct comparison when a useful inequality is easy.
- Use limit comparison when dominant behavior is clearer than exact inequalities.
- Use the integral test for positive, continuous, decreasing function-generated terms.
- Use the ratio test for factorials and strong exponential structure.
- Use the root test when an expression is raised to the \(n\)th power.
- For ratio and root tests, \(L=1\) means inconclusive—not divergence.
- Finite initial changes do not affect convergence.
- A convergence proof need not produce the exact sum.
Where We Go Next
Positive-term tests exploit the fact that all terms push partial sums in the same direction. Once signs begin to alternate, cancellation introduces a new mechanism that can create convergence even when the corresponding positive series diverges.
The next lesson, Alternating Series, develops the alternating series test, explains why decreasing terms approaching zero can converge through cancellation, and introduces the distinction between convergence caused by size alone and convergence caused by sign cancellation.