Why Group the Terms?
A p-series has terms that shrink as the index increases, but shrinking terms alone do not tell us whether their infinite sum is finite. To prove convergence for an exponent greater than one, we will group the terms into blocks whose lengths double. Within each block, every term has a simple upper bound, and the resulting block bounds form a geometric series.
This approach makes the role of the exponent visible. Each block contains more terms than the previous block, but the terms are small enough that the total bound for a block decreases geometrically.
The Dyadic Blocks
For each integer \(k\ge 0\), define the \(k\)th dyadic block to consist of the indices from \(2^k\) through \(2^{k+1}-1\), inclusive. The first few blocks are \(\{1\}\), \(\{2,3\}\), \(\{4,5,6,7\}\), and \(\{8,9,\ldots,15\}\).
These blocks do not overlap, and together they contain every positive integer. The \(k\)th block has \(2^{k+1}-2^k=2^k\) indices. Because \(n\ge 2^k\) throughout that block, and \(p>0\), each term satisfies \(\frac{1}{n^p}\le\frac{1}{(2^k)^p}\). Multiplying this bound by the number of terms gives an upper bound for the entire block.
The expression on the right is the key. It can be written as \((2^{1-p})^k\). If \(p>1\), then \(1-p<0\), so \(0<2^{1-p}<1\). Thus the upper bounds for the block totals are the terms of a convergent geometric series.
Why the Bound Proves Convergence
It is not enough to bound only partial sums that end exactly at the end of a block. A general partial sum may stop partway through a block. Positivity handles this: stopping partway through a block adds no more than including the whole block. Therefore every partial sum is bounded by a finite initial sum of the geometric block bounds.
Here is the argument in full. Let \(N\ge1\), and choose \(m\ge0\) so that \(2^m\le N<2^{m+1}\). The terms up to \(N\) lie in blocks \(0\) through \(m\), possibly using only part of block \(m\). Since all terms are positive, we may bound that partial block by its complete block total. We then apply the block estimate:
Because every term is positive, extending the last, possibly incomplete block can only increase the sum. Hence \(S_N\le\sum_{k=0}^{m}\sum_{n=2^k}^{2^{k+1}-1}\frac{1}{n^p}\).
For \(p>1\), this gives \(S_N\le\sum_{k=0}^{m}(2^{1-p})^k\).
Set \(q=2^{1-p}\), so \(0<q<1\). The geometric-series formula gives \(\sum_{k=0}^{m}q^k\le\frac{1}{1-q}\). Thus \(S_N\le\frac{1}{1-2^{1-p}}\) for every \(N\).
The partial sums increase because \(S_{N+1}-S_N=\frac{1}{(N+1)^p}>0\). They are bounded above by a finite constant. Every increasing sequence of real numbers that is bounded above converges, so the p-series converges.
The last step uses the monotone convergence property for real sequences. In this setting, it can be seen by taking \(L=\sup\{S_N:N\in\mathbb N\}\), which exists because the partial sums are bounded above. For any \(\varepsilon>0\), the definition of supremum gives some \(S_{N_0}>L-\varepsilon\). Since the partial sums increase, \(L-\varepsilon<S_N\le L\) whenever \(N\ge N_0\). Therefore \(S_N\) approaches the finite number \(L\).
Worked Examples
Worked Example: Bounding the Series with Exponent 2
Consider \(\sum_{n=1}^{\infty}\frac{1}{n^2}\). Here \(p=2\), so the block bound is \(2^{k(1-2)}=2^{-k}\).
The first block, \(k=0\), contains the term \(1\), and its total is \(1\le1\). The second block contains \(n=2,3\). Each term is at most \(1/2^2=1/4\), so its total is at most \(2(1/4)=1/2\). In fact, its total is \(1/4+1/9=13/36\le1/2\).
The third block contains \(n=4,5,6,7\). Each term is at most \(1/4^2=1/16\), so its total is at most \(4(1/16)=1/4\). The block bounds continue as \(1,\frac12,\frac14,\frac18,\ldots\), a geometric series with first term \(1\) and ratio \(1/2\).
Thus every partial sum of \(\sum_{n=1}^{\infty}\frac{1}{n^2}\) is at most \(2\), and the partial sums increase. The series converges. As a check on the block calculation, the bound for a block of \(2^k\) terms is \(2^k(2^k)^{-2}=2^{-k}\), exactly the bound used above.
Worked Example: Bounding the Series with Exponent \(3/2\)
Consider \(\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}\). Here \(p=3/2>1\). For block \(k\), there are \(2^k\) terms, and each is at most \((2^k)^{-3/2}\). Therefore the block total is at most \(2^k(2^k)^{-3/2}=2^{-k/2}=(1/\sqrt{2})^k\).
For example, block \(k=1\) contains \(n=2,3\), and its total satisfies \(\frac{1}{2^{3/2}}+\frac{1}{3^{3/2}}\le 2\cdot\frac{1}{2^{3/2}}=\frac{1}{\sqrt{2}}\). Block \(k=2\) contains four terms, each at most \(\frac{1}{4^{3/2}}=\frac18\), so its total is at most \(4\cdot\frac18=\frac12\). These agree with the bounds \((1/\sqrt2)^1\) and \((1/\sqrt2)^2\).
The geometric ratio is \(1/\sqrt2\), which is between zero and one. Its sum is \(\frac{1}{1-1/\sqrt2}=2+\sqrt2\), since multiplying numerator and denominator by \(\sqrt2\) gives \(\frac{\sqrt2}{\sqrt2-1}=\frac{\sqrt2(\sqrt2+1)}{2-1}=2+\sqrt2\). Consequently, all partial sums are at most \(2+\sqrt2\); they increase and are bounded, so this p-series converges. The value \(2+\sqrt2\) is an upper bound obtained from the block estimates, not an assertion that it equals the sum.
Worked Example: Bounding the Series with Exponent 3
For \(\sum_{n=1}^{\infty}\frac{1}{n^3}\), set \(p=3\). The bound for block \(k\) is \(2^{k(1-3)}=2^{-2k}=(1/4)^k\).
The first block contributes \(1\). In the second block, \(n=2,3\), each term is at most \(1/2^3=1/8\), so the block total is at most \(2/8=1/4\). In the third block, \(n=4,5,6,7\), each term is at most \(1/4^3=1/64\), so the total is at most \(4/64=1/16\). These bounds are \(1\), \(1/4\), and \(1/16\), the first three terms of the geometric sequence with ratio \(1/4\).
The sum of all block bounds is \(\frac{1}{1-1/4}=\frac43\). Thus the partial sums are bounded above by \(4/3\) and increase, proving convergence. As a check, the general block formula gives \(2^k(2^k)^{-3}=2^{-2k}\), the same bound found for each block.
A Common Pitfall: Counting the Block Correctly
A frequent error is to use the largest index in a block as though it gave an upper bound for every term. For instance, in block \(k=2\), the indices are \(4,5,6,7\). Since \(n\le7\), it is not true that \(1/n^p\le1/7^p\) for every index in that block. The term at \(n=4\) is larger than \(1/7^p\).
The correct upper bound uses the smallest index in the block: \(n\ge2^k\), so \(1/n^p\le1/(2^k)^p\). Then multiply by the correct number of indices, \(2^k\). This yields \(2^k/(2^k)^p=2^{k(1-p)}\). Keeping both the inequality direction and the block size straight is essential to the proof.
Why the Exponent Matters
The method succeeds when the block bounds decrease geometrically. The ratio is \(2^{1-p}\), and it is less than one precisely when \(p>1\). The proof therefore turns the exponent condition into a concrete fact about the geometric bound. If that ratio is not less than one, this particular geometric-series estimate cannot establish convergence; the convergence argument above relies on the strict inequality \(p>1\).
Grouping is useful because it controls many terms at once rather than requiring a separate estimate for each partial sum. The blocks partition the original series in its natural order, and positivity ensures that an incomplete final block never exceeds its complete-block bound. Together, these facts turn a local estimate on each block into a global bound on every partial sum.
Check Your Understanding
Use the dyadic-block argument to answer the following questions.
- How many indices are in the block from \(2^k\) to \(2^{k+1}-1\), inclusive?
- Why is each term in that block at most \(1/(2^k)^p\) when \(p>0\)?
- What is the geometric ratio of the block bounds for a p-series?
- Why does an upper bound for every complete block also help bound a partial sum that ends inside a block?
- Which two properties of the partial sums are used to conclude that the p-series converges when \(p>1\)?
