The p-Series

The Exponent Determines the Outcome

The Limit Comparison Test gives a way to compare positive series when their terms have a ratio approaching a positive finite number. To use it effectively, we need standard reference series whose convergence behavior is known. The p-series is one of the most important of these: its terms are simple powers, yet the exponent creates a sharp boundary between convergence and divergence.

We will prove that boundary by grouping terms into blocks whose endpoints are powers of two. Within each block, every term can be bounded using the same power of two, and the number of terms in the block is also a power of two. The resulting block estimates form a geometric series. This approach proves the classification without using the Integral Test, which will be introduced in the next tutorial.

Definition (p-Series): For a real number $p$, the series $$ \sum_{n=1}^{\infty}\frac{1}{n^p} $$ is called the p-series with exponent $p$. Its $n$th term is positive for every positive integer $n$.

The exponent $p$ may be positive, zero, or negative. We will treat all three possibilities. The main result is a complete classification: the series converges precisely when $p>1$.

Dyadic Blocks

For each integer $k\geq0$, form the block of indices from $2^k$ through $2^{k+1}-1$, inclusive. Each block contains exactly $2^k$ terms. The first few are

\{1\},\quad \{2,3\},\quad \{4,5,6,7\},\quad \{8,9,\ldots,15\}.

These blocks partition the positive integers: each positive integer belongs to exactly one block. When $p>0$, the function $x\mapsto x^{-p}$ decreases for $x>0$. Thus, in the $k$th block, the terms are no larger than the first term and no smaller than the value at the right endpoint, with strictness at the lower bound because the final index is less than $2^{k+1}$.

Block Bounds: For $p>0$ and $k\geq0$, $$ 2^{k(1-p)-p} \leq \sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p} \leq 2^{k(1-p)}. $$

To verify the bounds, there are $2^k$ terms in the block. For every such $n$, we have $2^k\leq n<2^{k+1}$. Since $p>0$,

2^{-(k+1)p}\leq \frac{1}{n^p}\leq 2^{-kp}.

Summing these inequalities over the $2^k$ indices gives

2^k2^{-(k+1)p} \leq \sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p} \leq 2^k2^{-kp}.

The left expression simplifies to $2^{k(1-p)-p}$, and the right expression simplifies to $2^{k(1-p)}$. These bounds show why $p=1$ is the threshold: the upper block estimate decreases geometrically when $p>1$, while the lower block estimate stays bounded away from zero when $0<p\leq1$.

The p-Series Classification

Theorem (p-Series Convergence Criterion): For real $p$, the p-series $$ \sum_{n=1}^{\infty}\frac{1}{n^p} $$ converges if and only if $p>1$.

Proof. First suppose $p>1$. The terms in each dyadic block are nonnegative, and the Block Bounds give

\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p} \leq 2^{k(1-p)}.

Put $q=2^{1-p}$. Since $p>1$, $0<q<1$, and $2^{k(1-p)}=q^k$. Summing the block upper bounds from $k=0$ through $k=K$ yields

\sum_{n=1}^{2^{K+1}-1}\frac{1}{n^p} \leq \sum_{k=0}^{K}q^k \leq \frac{1}{1-q}.

The last inequality follows from the finite geometric-sum formula, since $0<q<1$. Every partial sum of the p-series is nonnegative and is bounded above by $1/(1-q)$: for any index $N$, choose $K$ with $N\leq2^{K+1}-1$, and use that the partial sums are increasing. By the earlier Theorem (Bounded Increasing Partial Sums), the series converges.

Now suppose $0<p\leq1$. The lower Block Bound gives, for every $k\geq0$,

\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p} \geq 2^{k(1-p)-p} \geq 2^{-p}.

The second inequality holds because $1-p\geq0$, so $k(1-p)\geq0$. Adding the contributions from the first $K+1$ blocks gives

\sum_{n=1}^{2^{K+1}-1}\frac{1}{n^p} \geq (K+1)2^{-p}.

As $K$ increases, the right-hand side is unbounded. The partial sums are therefore unbounded, so the series diverges by the earlier Theorem (Unbounded Partial Sums Imply Divergence).

It remains to consider $p\leq0$. For every positive integer $n$, $n^{-p}\geq1$. In particular, the terms do not tend to zero. The earlier Necessary Condition for Series Convergence says that the terms of a convergent series must tend to zero, so the p-series diverges in this case as well. We have proved convergence for $p>1$ and divergence for every $p\leq1$. $\square$

Worked Applications

Worked Example: A Convergent p-Series

Consider

\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}.

Here $p=3/2>1$, so the p-Series Convergence Criterion gives convergence. The dyadic proof also supplies a bound on the contribution from each block. For the block beginning at $2^k$,

\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^{3/2}} \leq 2^{k(1-3/2)} =2^{-k/2}.

The block bounds sum to a geometric series with ratio $1/\sqrt{2}$, which is less than $1$. In particular, the blocks with $k=0,1,2$ contain the indices $1$, $2$ through $3$, and $4$ through $7$, respectively; their upper bounds are $1$, $1/\sqrt{2}$, and $1/2$. This illustrates how the block estimates control groups of terms even though they do not compute the exact sum.

Worked Example: The Boundary Exponent

The p-series with $p=1$ is

\sum_{n=1}^{\infty}\frac{1}{n}.

In every dyadic block, the lower Block Bound is

\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n} \geq 2^{k(1-1)-1} =\frac{1}{2}.

For example, the block from $2$ through $3$ contributes $1/2+1/3$, which is greater than $1/2$; the block from $4$ through $7$ contributes at least $4(1/8)=1/2$. After the first $K+1$ blocks, the partial sum is at least $(K+1)/2$, so the partial sums are unbounded and the harmonic series diverges. The same block argument works for every exponent $0<p<1$, with the lower bound $2^{-p}$ in place of $1/2$.

Worked Example: Comparing a Rational Series with a p-Series

Determine the behavior of

\sum_{n=1}^{\infty}\frac{3n+2}{n^3+5n+1}.

The numerator and denominator are positive for $n\geq1$, so the terms are positive. Compare with $b_n=1/n^2$. The ratio is

\frac{\dfrac{3n+2}{n^3+5n+1}}{1/n^2} = \frac{3n^3+2n^2}{n^3+5n+1} = \frac{3+2/n}{1+5/n^2+1/n^3} \longrightarrow 3.

The limit is positive and finite. By the Limit Comparison Test, the rational series and $\sum 1/n^2$ have the same convergence behavior. Since $2>1$, the p-Series Convergence Criterion shows that $\sum 1/n^2$ converges. Therefore the rational series converges as well.

A Tail Estimate from the Same Blocks

For $p>1$, the dyadic argument gives more than convergence. It also bounds the remainder after a partial sum. If $S_N=\sum_{n=1}^{N}1/n^p$, write

R_N=\sum_{n=N+1}^{\infty}\frac{1}{n^p}.

The remainder is a convergent tail and is positive. A particularly direct estimate is available when the tail begins at a power of two.

Theorem (Dyadic Tail Bound for a p-Series): If $p>1$ and $m\geq0$ is an integer, then $$ \sum_{n=2^m}^{\infty}\frac{1}{n^p} \leq \frac{2^{m(1-p)}}{1-2^{1-p}}. $$

Proof. The tail starting at $2^m$ is the union of the dyadic blocks indexed by $k=m,m+1,\ldots$. By the upper Block Bound, each such block contributes at most $2^{k(1-p)}$. Therefore every finite sum of these blocks satisfies

\sum_{k=m}^{K}\ \sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p} \leq \sum_{k=m}^{K}2^{k(1-p)}.

Let $q=2^{1-p}$, so $0<q<1$. The finite geometric sum on the right is at most $q^m/(1-q)$. As $K$ tends to infinity, the left side increases to the full tail, and the same upper bound remains valid. Since $q^m=2^{m(1-p)}$, this proves the stated estimate. $\square$

For a general $N\geq0$, choose $m\geq0$ so that $2^m\leq N+1<2^{m+1}$. The remainder after $N$ is no larger than the tail beginning at $2^m$. Also $2^m>(N+1)/2$, and $1-p<0$, so

R_N \leq \frac{2^{m(1-p)}}{1-2^{1-p}} < \frac{2^{p-1}(N+1)^{1-p}}{1-2^{1-p}}.

Thus the remainder is bounded by a constant times $(N+1)^{1-p}$. This estimate is useful for controlling how much of a convergent p-series remains after a finite partial sum. It is an upper bound, not an exact formula for the remainder.

Why the Block Method Matters

The p-series classification is a practical benchmark for comparison tests. If a positive term sequence has the same size, up to a positive finite ratio, as $1/n^p$, the Limit Comparison Test transfers the p-series verdict to the new series. For instance, the rational series in the worked example inherits convergence from $1/n^2$.

The block proof also explains the threshold rather than merely stating it. When $p>1$, the total contribution of each successive block is bounded by a geometric sequence that shrinks. When $0<p\leq1$, every block contributes at least a fixed positive amount, so infinitely many blocks force unbounded partial sums. A common error is to look only at the fact that $1/n^p\to0$ for $p>0$ and conclude convergence. The Necessary Condition for Series Convergence is only a necessary condition: terms tending to zero do not guarantee that a series converges. The block lower bounds show precisely why that reasoning fails at and below the boundary exponent.

Identify the exponent.
Write the comparison series as $1/n^p$ and determine whether $p$ is above, equal to, or below $1$.

Use the classification.
For $p>1$ the p-series converges; for $p\leq1$ it diverges.

Transfer the conclusion when needed.
For another eventually positive series, use a direct comparison or the Limit Comparison Test with the appropriate p-series.

Takeaway: The p-series $\sum 1/n^p$ converges exactly when $p>1$. Dyadic blocks prove both sides of this threshold and give a geometric upper bound for tails when $p>1$.

Check Your Understanding

Use the dyadic block estimates and the p-series classification to answer each question.

How many terms are in the block of indices from $2^k$ through $2^{k+1}-1$?
Why does the upper block estimate produce a convergent geometric series when $p>1$?
For $0<p\leq1$, what lower bound does each dyadic block contribute?
Why does the p-series diverge when $p\leq0$, without using the block estimates?
For $p>1$, what additional information does the Dyadic Tail Bound provide beyond the fact that the series converges?

What You'll Learn