Two Familiar Series Patterns
A series can be easier to classify once its terms are recognized as belonging to a familiar pattern. Two important patterns are geometric series, whose terms have a fixed multiplicative ratio, and p-series, whose terms follow a power law.
These patterns can look similar. In particular, the ratio of consecutive terms in a p-series approaches \(1\) when the exponent is nonzero. But approaching a fixed number is not the same as being equal to that number at every step. That difference is essential: the geometric-series rule and the p-series rule are different.
Definitions and Convergence Rules
A geometric series has terms of the form \(a r^{n-1}\), where \(a\) is its first term and \(r\) is its common ratio. For nonzero consecutive terms, dividing a term by the preceding term gives the same value \(r\) each time. A geometric series with \(a\ne 0\) converges exactly when \(|r|<1\); if \(|r|\ge 1\), it diverges.
A p-series has terms \(1/n^p\), where \(p\) is a real number. A nonzero constant multiplier does not change whether it converges. The p-series converges when \(p>1\) and diverges when \(p\le 1\). In particular, \(p=0\) gives constant terms, not a decaying power law.
A shifted p-series has terms such as \(C/(n+s)^p\), with a fixed shift \(s\) for which the denominators in the series are positive, and a fixed nonzero multiplier \(C\). Shifting the starting index or adding a fixed shift to the index does not change the p-series threshold: its classification is still determined by \(p>1\) versus \(p\le 1\).
Using Consecutive-Term Ratios Carefully
For a sequence of nonzero terms \(a_n\), its consecutive-term ratio is \(a_{n+1}/a_n\). If this ratio has the same value for every \(n\), the terms follow a geometric pattern. If the ratio depends on \(n\), the terms are not a geometric sequence with a fixed nonzero ratio.
For p-series terms \(a_n=1/n^p\), calculate the ratio directly:
When \(p\ne 0\), this ratio varies with \(n\). Indeed, \(n/(n+1)\) increases strictly as \(n\) increases, and raising positive numbers to a fixed nonzero power preserves strict monotonicity when \(p>0\) and reverses it when \(p<0\). The ratio also tends to \(1\). For \(p=0\), however, the ratio is exactly \(1\) for every \(n\), because every term equals \(1\). That case is geometric, with ratio \(1\), and the series diverges.
The same reasoning applies to shifted terms \(C/(n+s)^p\), when the terms are defined and \(C\ne 0\). Their consecutive-term ratio is \(((n+s)/(n+s+1))^p\). For \(p\ne0\), this depends on \(n\); for \(p=0\), the terms are constant and the ratio is \(1\).
Why the Ratio Distinguishes the Patterns
The distinction follows from how each pattern is built. In a geometric sequence, each new term is obtained by multiplying the preceding term by one fixed number. Dividing the new term by the preceding term therefore returns that same number at every step. In contrast, a p-series term is determined by the index raised to a power, so its ratio compares two different index values.
If \(a_n=a r^{n-1}\), then \(a_{n+1}=a r^n\). Provided the terms being divided are nonzero, \(a_{n+1}/a_n=r\), independent of \(n\).
If \(a_n=1/n^p\), then division gives \(a_{n+1}/a_n=(n/(n+1))^p\), which depends on \(n\) when \(p\ne0\).
A fixed geometric ratio calls for the geometric-series rule. A power-law form calls for the p-series rule, including its \(p=0\) case.
This is a recognition method, not a general convergence test. A ratio that tends to a number does not by itself make a series geometric. For example, p-series ratios with \(p\ne0\) tend to \(1\), although their convergence depends on whether \(p\) is greater than \(1\).
Worked Examples
Worked Example: A Geometric Series with Ratio Between Zero and One
Classify \(\sum_{n=1}^{\infty}6(1/3)^{n-1}\). Its first terms are \(6\), \(2\), \(2/3\), and \(2/9\). Dividing each term by the preceding term gives \(2/6=1/3\), \((2/3)/2=1/3\), and \((2/9)/(2/3)=1/3\). Thus the series is geometric with \(a=6\) and \(r=1/3\).
Since \(|r|=1/3<1\), the geometric series converges. Its sum is
Check the result using the finite partial sum. The first \(N\) terms sum to \(6(1-(1/3)^N)/(1-1/3)=9(1-(1/3)^N)\). As \(N\) increases, \((1/3)^N\) tends to \(0\), so these partial sums tend to \(9\), as claimed.
Worked Example: A Shifted p-Series with Exponent Greater Than One
Classify \(\sum_{n=1}^{\infty}5/(n+2)^2\). Its first terms are \(5/9\), \(5/16\), and \(5/25\). The ratio of the second term to the first is \((5/16)/(5/9)=9/16\); the ratio of the third to the second is \((5/25)/(5/16)=16/25\). These ratios differ, so the terms do not have a fixed geometric ratio.
The terms have the shifted power-law form \(5/(n+2)^2\), with exponent \(p=2\) and a fixed multiplier \(5\). Because \(2>1\), the shifted p-series rule says the series converges.
Check the classification against the pattern: removing the multiplier \(5\) and the fixed shift \(2\) leaves a p-series with exponent \(2\). Neither the shift nor the nonzero multiplier changes the convergence classification, so convergence is the appropriate conclusion.
Worked Example: A Shifted Harmonic Series
Classify \(\sum_{n=1}^{\infty}4/(n+1)\). The first three terms are \(4/2=2\), \(4/3\), and \(4/4=1\). The first consecutive-term ratio is \((4/3)/2=2/3\), while the next is \(1/(4/3)=3/4\). The ratios are not equal, so this is not a geometric series.
The terms are \(4/(n+1)^1\), a shifted p-series with exponent \(p=1\) and multiplier \(4\). Since \(1\le1\), the p-series rule gives divergence.
Check: the exponent is exactly the boundary value \(1\), and the rule includes \(p=1\) among the divergent cases. The shift and multiplier do not move the series across that boundary.
Worked Example: The Constant-Term Case \(p=0\)
Consider \(\sum_{n=1}^{\infty}7/n^0\). Since \(n^0=1\) for every positive integer \(n\), each term is \(7\). The first terms are \(7,7,7\), and the consecutive-term ratios are \(7/7=1\) and \(7/7=1\). The ratio is fixed, so this is also a geometric series with ratio \(1\).
The p-series rule gives divergence because \(p=0\le1\). The geometric rule agrees: \(|r|=1\), not less than \(1\), so the geometric series diverges.
Check with partial sums: the sum of the first \(N\) terms is \(7N\), which grows without bound as \(N\) increases. Thus the series diverges, consistent with both rules.
A Common Mistake: Treating a Limiting Ratio as Fixed
A tempting but incorrect argument is to compute a ratio, take its limit, and then treat that limit as the common ratio of a geometric series. Consider \(\sum_{n=1}^{\infty}1/n^2\). Its consecutive-term ratio is
For \(n=1\), this ratio is \(1/4\); for \(n=2\), it is \(4/9\). The ratios are not fixed, although they approach \(1\). It is therefore wrong to call this a geometric series with ratio \(1\), and wrong to conclude divergence from the geometric-series rule.
The correction is to recognize the actual term pattern: \(1/n^2\) is a p-series with \(p=2\). Since \(2>1\), it converges. The limiting ratio is not a substitute for identifying the series.
Why the Distinction Matters
The two patterns answer different questions. A geometric series is classified by the size of its fixed ratio: ratios with absolute value below \(1\) produce convergence, while ratios with absolute value at least \(1\) produce divergence. A p-series is classified by its exponent: it converges only when \(p>1\). The p-series with \(p=0\) is a special overlap in appearance, since its terms are constant and are also geometric with ratio \(1\); both rules give divergence.
Some series may fit neither pattern. For instance, a term involving both an index factor and an exponential factor need not be geometric or a p-series. In such cases, do not force either classification. Other methods are needed, and recognizing that a pattern does not apply is part of correct classification.
Check Your Understanding
For each question, identify the pattern before deciding which convergence rule applies.
- For a sequence of nonzero terms, what feature of its consecutive-term ratios identifies a geometric pattern?
- What is the consecutive-term ratio for \(a_n=1/n^3\), and does it stay fixed?
- Classify \(\sum_{n=1}^{\infty}2/(n+4)^3\). State the exponent and the relevant rule.
- Why does \(\sum_{n=1}^{\infty}1/n^2\) not become geometric merely because its consecutive-term ratios tend to \(1\)?
- What pattern do the terms \(9/n^0\) form, and does the resulting series converge?
