Why p-Series Diverge When the Exponent Is at Most One

Why the Exponent Threshold Matters

A p-series has terms that get smaller as the index grows, provided its exponent is positive. But smaller terms do not by themselves guarantee that their infinite sum is finite. The key question is how quickly the terms shrink compared with how many terms accumulate.

For exponents $0<p\leq1$, a useful way to answer that question is to group terms into blocks whose lengths double: first one term, then two, then four, then eight, and so on. Within each block, every term has a lower bound determined by the block’s largest possible index. The resulting block sums stay above a fixed positive amount, so the partial sums cannot remain bounded.

Definition: For a real exponent $p>0$, the p-series is $\sum_{n=1}^{\infty}1/n^p$. Its $N$th partial sum is $S_N=\sum_{n=1}^{N}1/n^p$. The series converges if the sequence of partial sums has a finite limit; if its partial sums are unbounded, the series diverges.

Build the Dyadic Blocks

For each integer $k\geq0$, use the block of indices from $2^k$ through $2^{k+1}-1$, inclusive. The block contains $2^k$ terms: the number of integers in it is

(2^{k+1}-1)-2^k+1=2^{k+1}-2^k=2^k.

Every index $n$ in this block satisfies $n<2^{k+1}$. Since $p>0$, taking reciprocal $p$th powers reverses this comparison, giving

\frac{1}{n^p}>\frac{1}{(2^{k+1})^p}.

There are $2^k$ terms, and each is greater than this same lower bound. Thus the total contribution of the block is greater than

2^k\frac{1}{(2^{k+1})^p} =2^k\,2^{-p(k+1)} =2^{-p}2^{k(1-p)}.

This expression is the central estimate. When $p\leq1$, the exponent $1-p$ is nonnegative. Therefore $2^{k(1-p)}\geq1$ for every $k\geq0$, and each block contributes more than $2^{-p}$. The precise block contribution may increase as $k$ grows, but it never drops below this fixed positive amount.

Key estimate: If $0<p\leq1$, then for every integer $k\geq0$, $$ \sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p} >2^{-p}2^{k(1-p)} \geq2^{-p}. $$

Why a Fixed Block Contribution Forces Divergence

The first $K$ dyadic blocks, for a positive integer $K$, cover the indices from $1$ through $2^K-1$. Adding the estimate for each block gives

S_{2^K-1} =\sum_{k=0}^{K-1}\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p} >\sum_{k=0}^{K-1}2^{-p} =K2^{-p}.

For a fixed $p$, $2^{-p}$ is a positive constant. As $K$ increases, $K2^{-p}$ grows without bound. The partial sums along the endpoints $2^K-1$ are therefore unbounded. If all partial sums had a finite limit, they would be bounded; hence the series cannot converge.

Theorem: For every real exponent $p$ with $0<p\leq1$, the p-series $\sum_{n=1}^{\infty}1/n^p$ diverges. The proof uses dyadic blocks, each of which contributes more than the same positive amount $2^{-p}$.

Worked Examples

Worked Example: The Harmonic Series

Consider $p=1$, so the terms are $1/n$. For this exponent, the block estimate becomes

\sum_{n=2^k}^{2^{k+1}-1}\frac1n >2^{-1}2^{k(1-1)} =\frac12\cdot1 =\frac12.

The first four blocks are $n=1$, $n=2,3$, $n=4,5,6,7$, and $n=8,\ldots,15$. Check their contributions one at a time:

\begin{aligned} B_0&=1>\frac12,\\ B_1&=\frac12+\frac13>\frac14+\frac14=\frac12,\\ B_2&=\frac14+\frac15+\frac16+\frac17>4\left(\frac18\right)=\frac12,\\ B_3&=\sum_{n=8}^{15}\frac1n>8\left(\frac1{16}\right)=\frac12. \end{aligned}

The strict lower bounds for $B_1,B_2,B_3$ follow because each index in those blocks is less than the next power of two. Adding the four displayed block estimates gives

S_{15}=B_0+B_1+B_2+B_3 >1+\frac12+\frac12+\frac12 =\frac52.

Check the block counts: the four blocks have $1,2,4,8$ terms, respectively, and together cover $1$ through $15$. The general estimate gives more than $K/2$ through the endpoint $2^K-1$; setting $K=4$ gives $S_{15}>2$, which is consistent with the sharper block-by-block bound $S_{15}>5/2$. As $K$ grows, $K/2$ has no finite bound, so the harmonic series diverges.

Worked Example: Exponent One-Half

Now take $p=\frac12$. For block $k$, the estimate is

\sum_{n=2^k}^{2^{k+1}-1}\frac1{\sqrt n} >2^{-1/2}2^{k/2}.

Compute the first four block bounds, showing the indices and the common lower bound for terms in each block:

\begin{aligned} B_0&=1>\frac1{\sqrt2},\\ B_1&=\frac1{\sqrt2}+\frac1{\sqrt3} >2\left(\frac1{\sqrt4}\right)=1,\\ B_2&=\sum_{n=4}^{7}\frac1{\sqrt n} >4\left(\frac1{\sqrt8}\right) =\frac4{2\sqrt2} =\sqrt2,\\ B_3&=\sum_{n=8}^{15}\frac1{\sqrt n} >8\left(\frac1{\sqrt{16}}\right) =8\left(\frac14\right)=2. \end{aligned}

For example, every index from $4$ to $7$ is less than $8$, so each corresponding term is greater than $1/\sqrt8$; there are four such terms. Likewise, the block from $8$ to $15$ has eight terms, each greater than $1/\sqrt{16}=1/4$. Adding the bounds gives

S_{15}=B_0+B_1+B_2+B_3>1+1+\sqrt2+2=4+\sqrt2.

Check the general block formula at $k=2$: it gives $2^{-1/2}2^{2/2}=2^{-1/2}\cdot2=\sqrt2$, exactly the bound obtained for $B_2$. The first four block bounds are consistent with the formula, and every additional block contributes more than $2^{-1/2}$. Consequently the partial sums are unbounded and this p-series diverges.

Worked Example: Exponent Three-Quarters

Let $p=\frac34$. The block estimate becomes

B_k=\sum_{n=2^k}^{2^{k+1}-1}\frac1{n^{3/4}} >2^{-3/4}2^{k/4}.

For $k=0,1,2,3$, calculate each bound:

\begin{aligned} B_0&=1>2^{-3/4},\\ B_1&=\sum_{n=2}^{3}\frac1{n^{3/4}} >2\left(\frac1{4^{3/4}}\right) =2\left(\frac1{2\sqrt2}\right) =2^{-1/2},\\ B_2&=\sum_{n=4}^{7}\frac1{n^{3/4}} >4\left(\frac1{8^{3/4}}\right) =4\left(2^{-9/4}\right) =2^{-1/4},\\ B_3&=\sum_{n=8}^{15}\frac1{n^{3/4}} >8\left(\frac1{16^{3/4}}\right) =8\left(\frac18\right) =1. \end{aligned}

The endpoint calculations use $4^{3/4}=2\sqrt2$, $8^{3/4}=2^{9/4}$, and $16^{3/4}=8$. The block sizes are $1,2,4,8$, matching the multipliers used in each line. Therefore the sum through $15$ satisfies

S_{15}=B_0+B_1+B_2+B_3 >2^{-3/4}+2^{-1/2}+2^{-1/4}+1.

As a check, substituting $k=3$ into the general estimate gives $2^{-3/4}2^{3/4}=1$, the bound calculated for the fourth block. For every later block, the bound $2^{-3/4}2^{k/4}$ remains at least $2^{-3/4}$. Thus arbitrarily many blocks add at least a fixed positive amount each, and the series diverges.

What Changes at the Boundary Exponent?

The same proof covers both $0<p<1$ and $p=1$, but the block estimates behave slightly differently. If $p=1$, then $k(1-p)=0$, so the estimate for every block is greater than $1/2$. If $0<p<1$, then $k(1-p)$ increases with $k$, and the lower bound $2^{-p}2^{k(1-p)}$ increases as the blocks move farther out.

The proof needs only that each block contributes at least a fixed positive amount; it does not need the amount to increase. This is why the boundary case $p=1$ is included. The argument also depends on $p>0$, which ensures that $1/n^p$ is a decreasing function of the positive integer index and justifies the endpoint bound used within each block.

Common Pitfall: Reversing the Useful Bound

A tempting but unhelpful step is to use the lower endpoint $n\geq2^k$ to claim a lower bound on each term. For $p>0$, that inequality actually gives an upper bound:

n\geq2^k \quad\Longrightarrow\quad \frac1{n^p}\leq\frac1{(2^k)^p}.

For example, in the block $n=4,5,6,7$ with $p=1$, this reasoning gives $1/n\leq1/4$. It is incorrect to then conclude that the block sum is at least $4(1/4)=1$: the terms $1/5,1/6,1/7$ are each less than $1/4$. This calculation supplies an upper bound on the block, not the lower bound needed to prove divergence.

The correction is to use the other end of the block. Since $n\leq7<8$, every term in that block satisfies $1/n>1/8$, so

\sum_{n=4}^{7}\frac1n>4\left(\frac18\right)=\frac12.

In general, use $n<2^{k+1}$ to obtain a lower bound for every term in block $k$, then multiply that lower bound by the correct number of terms, $2^k$.

Why the Block Argument Is Useful

A term-by-term glance can be misleading: the terms of these p-series approach zero, yet their cumulative sums do not settle at a finite value. The dyadic grouping makes the accumulation visible. Each time the scale doubles, there are enough terms in the new block to offset the decrease in the size of individual terms.

The proof is also economical. Rather than calculating the exact value of every partial sum, it selects a sequence of endpoints and proves a lower bound that grows without limit. The specific endpoint sequence $2^K-1$ is useful because it consists exactly of the first $K$ complete dyadic blocks. The conclusion follows from the block estimate and the definition of convergence, without needing to evaluate the infinite sum.

Check Your Understanding

Answer these questions using the block estimate and the definitions above.

How many integers are in the block from $2^k$ through $2^{k+1}-1$?
Why does $n<2^{k+1}$ give a lower bound for $1/n^p$ when $p>0$?
What lower bound does the dyadic-block argument give when $p=1$?
For $p=\frac12$, calculate the general lower bound for block $k=2$.
Why do unbounded partial sums along the endpoints $2^K-1$ rule out convergence?