Two Series with Efficient Classification Rules
A geometric series and a p-series are two standard families for which a convergence decision can be made from a single parameter. For a geometric series, that parameter is the common ratio. For a p-series, it is the exponent. The key is to state the correct condition and connect it to the conclusion; recognizing a familiar pattern without naming its condition is not a complete justification.
A series \(\sum_{n=1}^{\infty}a_n\) converges when its sequence of partial sums \(S_N=\sum_{n=1}^{N}a_n\) approaches a finite limit as \(N\) increases. If the partial sums do not approach a finite limit, the series diverges.
Classifying a Geometric Series
Write a geometric series with first term \(a\) and common ratio \(r\) as \(\sum_{n=1}^{\infty} ar^{n-1}\). Its terms begin \(a, ar, ar^2,\ldots\). If \(a=0\), every term is zero and the series converges to zero. Otherwise, classify it by the magnitude of \(r\).
The condition is strict: a ratio of magnitude exactly one does not give convergence. Here is why the rule works. The \(N\)th partial sum is \(S_N=a+ar+\cdots+ar^{N-1}\). For \(r\ne1\), multiply this equation by \(r\) and subtract:
When \(|r|<1\), the powers \(r^N\) approach zero. Thus \(S_N\) approaches \(\frac{a}{1-r}\), a finite limit. If \(|r|>1\), the terms \(ar^{N-1}\) do not approach zero, so the series cannot converge. If \(r=1\), all terms equal \(a\), and the partial sums are \(Na\), which do not approach a finite limit for \(a\ne0\). If \(r=-1\), the partial sums alternate between \(a\) and \(0\), so they also have no finite limit. These cases establish the classification for real \(r\).
Worked Example: Finding the Sum of a Convergent Geometric Series
Classify \(\sum_{n=1}^{\infty}6\left(-\frac{2}{3}\right)^{n-1}\), and find its sum if it converges.
The first term is \(a=6\). Each term is obtained by multiplying the preceding term by \(r=-\frac{2}{3}\), so the common ratio is \(r=-\frac{2}{3}\).
Its magnitude is \(|r|=\frac{2}{3}<1\). Therefore the geometric-series theorem guarantees convergence. The sum is
Check the result using the partial-sum formula: \(S_N=\frac{6(1-(-\frac{2}{3})^N)}{1+\frac{2}{3}}=\frac{18}{5}\left(1-(-\frac{2}{3})^N\right)\). Since \((-\frac{2}{3})^N\) approaches zero, \(S_N\) approaches \(\frac{18}{5}\), as claimed.
Worked Example: A Geometric Series with Ratio of Magnitude One
Classify \(\sum_{n=1}^{\infty}5(-1)^{n-1}\).
The first term is \(5\), and the common ratio is \(r=-1\). Its magnitude is \(|r|=1\), which does not satisfy the required condition \(|r|<1\). More directly, the partial sums begin \(5, 0, 5, 0,\ldots\), so they do not approach one finite limit. The series diverges.
This example shows why the condition cannot be weakened to \(|r|\le1\).
Classifying a p-Series
For a p-series, the classification depends on the exponent \(p\). The convergent range is \(p>1\), not merely \(p>0\). For \(p\le0\), the terms do not approach zero; for \(0<p\le1\), the terms do approach zero, but their sum still diverges.
Grouping consecutive terms into blocks whose sizes double proves both parts of the classification. Block \(k\), for \(k=0,1,2,\ldots\), consists of terms with indices \(2^k\le n\le2^{k+1}-1\). It contains \(2^k\) terms. Because all terms are positive, lower or upper bounds on each block can be used to establish the behavior of the partial sums.
Every index in block \(k\) satisfies \(n\ge2^k\), so \(\frac{1}{n^p}\le\frac{1}{(2^k)^p}\). The block sum is at most \(2^k\cdot2^{-kp}=2^{-k(p-1)}\).
The block bounds have ratio \(2^{-(p-1)}\), which is less than one when \(p>1\). Their infinite sum is \(\sum_{k=0}^{\infty}2^{-k(p-1)}=\frac{1}{1-2^{-(p-1)}}\), a finite number. Thus the partial sums of the p-series are increasing and bounded above, so they converge.
Every index in block \(k\) is less than \(2^{k+1}\), so \(\frac{1}{n^p}>\frac{1}{(2^{k+1})^p}\). The block sum is greater than \(2^k\cdot2^{-p(k+1)}=2^{-p}2^{k(1-p)}\ge2^{-p}\).
Infinitely many blocks each contribute more than the fixed positive amount \(2^{-p}\), so their partial sums cannot have a finite limit. If \(p\le0\), then \(n^{-p}\ge1\) for every \(n\), so the terms do not approach zero. The series diverges in this range as well.
The proof identifies an important distinction. A term approaching zero is necessary for a series to converge, but for \(0<p\le1\) it is not enough: the grouped terms still make the partial sums grow without bound.
Worked Example: Justifying Convergence of a p-Series
Classify \(\sum_{n=1}^{\infty}\frac{1}{n^{5/3}}\).
This is a p-series with \(p=\frac{5}{3}\). Since \(\frac{5}{3}>1\), the p-series theorem gives convergence. To see the block bound for this exponent, block \(k\) has \(2^k\) terms, and each term is at most \(\frac{1}{(2^k)^{5/3}}=2^{-5k/3}\). Therefore its sum is at most
The bounds across the blocks form a geometric series with first term \(1\) and ratio \(2^{-2/3}\). Since \(0<2^{-2/3}<1\), their sum is \(\frac{1}{1-2^{-2/3}}\), which is finite. The original series is positive and its block sums are bounded above by this finite total, so its partial sums converge. The exponent satisfies the required condition, and the series converges.
Worked Example: Justifying Divergence of the Harmonic Series
Classify the harmonic series \(\sum_{n=1}^{\infty}\frac{1}{n}\).
This is a p-series with \(p=1\). Since \(p\le1\), the p-series theorem says it diverges. The block argument makes the reason explicit. In block \(k\), there are \(2^k\) terms, and every index \(n\) is less than \(2^{k+1}\). Hence each term is greater than \(\frac{1}{2^{k+1}}\), and the block sum is greater than
For example, the block from \(n=1\) through \(n=1\) contributes \(1\); the block from \(n=2\) through \(n=3\) contributes \(\frac{1}{2}+\frac{1}{3}>\frac{1}{2}\); the block from \(n=4\) through \(n=7\) contributes \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>\frac{1}{2}\). Every further block also contributes more than \(\frac{1}{2}\). Adding infinitely many such blocks forces the partial sums to grow without bound, so the series diverges.
Writing a Complete Classification
For an AP justification, name the series type, identify its parameter, state the applicable condition, and connect that condition to the conclusion. When claiming convergence for a geometric series, you can also give the sum if requested. When claiming divergence, use the relevant classification condition rather than suggesting that a few terms demonstrate the outcome.
A Common Mistake: Reversing the p-Series Condition
A frequent error is to see the exponent \(p\) in the denominator and conclude that smaller exponents make the terms decrease more slowly, so the series should converge. For example, this reasoning might claim that \(\sum_{n=1}^{\infty}\frac{1}{n^{1/2}}\) converges because \(1/2>0\). That is incorrect: the convergence condition is \(p>1\), not \(p>0\).
For \(p=\frac12\), block \(k\) contains \(2^k\) terms, each greater than \(\frac{1}{(2^{k+1})^{1/2}}\). Therefore its block sum is greater than
These lower bounds do not approach zero; they grow as \(k\) increases. In particular, every block contributes at least a positive amount, and the series diverges. The correction is to check the precise threshold: \(\frac12\le1\), so the p-series diverges.
Check Your Understanding
For each question, identify the relevant parameter and justify the conclusion using the applicable classification rule.
- Does \(\sum_{n=1}^{\infty}7\left(\frac{1}{4}\right)^{n-1}\) converge? If so, what is its sum?
- Why does a geometric series with common ratio \(r=\frac{3}{2}\) diverge?
- Does \(\sum_{n=1}^{\infty}\frac{1}{n^{7/4}}\) converge or diverge?
- Does \(\sum_{n=1}^{\infty}\frac{1}{n^{2/3}}\) converge or diverge? Give the exponent condition.
- Why is \(|r|=1\) not enough for a nonzero geometric series to converge?
