The Cauchy Criterion for Series

Controlling Every Late Block

The Cauchy Criterion for Series, established earlier in this course, gives a way to decide convergence without first finding a proposed sum. Its key feature is that it tests finite blocks of terms: once the starting index is sufficiently large, every consecutive finite sum must be small. This is stronger than checking only a particular collection of blocks, such as pairs or blocks whose lengths have been fixed in advance.

Let $S_N=\sum_{n=1}^{N}a_n$, with $S_0=0$. For integers $q\geq p\geq1$, the Block-Sum Identity gives $$ \sum_{n=p}^{q}a_n=S_q-S_{p-1}. $$ Thus the Cauchy condition on partial sums is exactly a condition on all finite blocks of terms. Recall the result by its established name; it will serve as a tool here, not as a new theorem.

Recall (Cauchy Criterion for Series): The series $\sum_{n=1}^{\infty}a_n$ converges if and only if, for every $\varepsilon>0$, there is an integer $N$ such that $$ \left|\sum_{n=p}^{q}a_n\right|<\varepsilon $$ whenever $q\geq p\geq N$.

The quantifiers matter. The same $N$ must work for every later starting index $p$ and every finite endpoint $q\geq p$. It is not enough for one selected family of blocks to have small sums, or for the terms $a_n$ individually to be small. The condition asks whether cancellation and size together keep all finite sums far out in the series under control.

A Direct Test for Failure

To show that a series fails the Cauchy condition, it is enough to find one positive number $\varepsilon_0$ such that, no matter how far out one starts, there is a finite block whose sum has magnitude at least $\varepsilon_0$. The block may depend on the proposed starting index. This is often an efficient way to prove divergence: choose a block with a convenient length and estimate all its terms in the same direction.

Worked Example: Detecting Divergence with Blocks in the Harmonic Series

Consider $\sum_{n=1}^{\infty}1/n$. Given any integer $N$, choose an integer $m\geq N$, and take the block from $m+1$ through $2m$. It contains exactly $m$ terms. For each index $k$ in this block, $k\leq2m$, so $$ \frac{1}{k}\geq\frac{1}{2m}. $$ Adding these $m$ inequalities gives $$ \sum_{k=m+1}^{2m}\frac{1}{k}\geq m\cdot\frac{1}{2m}=\frac{1}{2}. $$ Thus, however large $N$ is, there is a block starting at an index at least $N$ whose sum is at least $1/2$. The Cauchy condition fails with $\varepsilon_0=1/2$, so the series diverges.

This argument uses neither a proposed value for the sum nor a comparison of partial sums with a limit. Its essential step is finding a late block whose total cannot be made small.

The same logic applies to signed series, but signs make the block choice more delicate. A large number of terms need not give a large block sum if positive and negative terms cancel. The next result makes this point precise for a useful kind of block decomposition.

A Criterion for Consecutive Blocks

Partition the positive integers into consecutive, nonempty, finite blocks. More precisely, let $$ 0=m_0<m_1<m_2<\cdots $$ be integers with $m_k\to\infty$, and let block $k$ contain the indices $m_{k-1}+1,\ldots,m_k$. Write its total as $$ B_k=\sum_{n=m_{k-1}+1}^{m_k}a_n. $$ A block total records the net result across the whole block. It does not record how far the partial sums may move before reaching that total. Define the within-block excursion $$ d_k=\max_{m_{k-1}<r\leq m_k} \left|\sum_{n=m_{k-1}+1}^{r}a_n\right|. $$ This maximum exists because block $k$ has only finitely many indices.

Theorem (Block Criterion with Within-Block Control): With the consecutive blocks defined above, $\sum_{n=1}^{\infty}a_n$ converges if and only if both the grouped series $\sum_{k=1}^{\infty}B_k$ converges and $d_k\to0$.

Proof. First suppose $\sum_{n=1}^{\infty}a_n$ converges. Its partial sums $S_N$ then converge. The partial sums of the grouped series are $$ \sum_{k=1}^{K}B_k =\sum_{n=1}^{m_K}a_n =S_{m_K}. $$ Since $m_K\to\infty$, these are a subsequence of the convergent sequence $(S_N)$, so they converge. Therefore $\sum B_k$ converges.

It remains to show $d_k\to0$. Fix $\varepsilon>0$. By the Cauchy Criterion for Series, there is an $N$ such that every finite block starting at an index at least $N$ has sum of magnitude less than $\varepsilon$. Choose $K$ so large that $m_{K-1}+1\geq N$. For $k\geq K$ and every $r$ in block $k$, the sum $$ \sum_{n=m_{k-1}+1}^{r}a_n $$ is one of the finite blocks covered by that criterion. Its magnitude is less than $\varepsilon$, so $d_k\leq\varepsilon$ for every $k\geq K$. Applying this argument with any smaller positive tolerance shows $d_k\to0$.

Conversely, suppose $\sum B_k$ converges and $d_k\to0$. Fix $\varepsilon>0$. The Cauchy Criterion for Series, applied to the grouped series, gives an integer $K_1$ such that $$ \left|\sum_{k=u}^{v}B_k\right|<\frac{\varepsilon}{2} $$ whenever $v\geq u\geq K_1$. Since $d_k\to0$, there is an integer $K_2$ such that $d_k<\varepsilon/4$ for every $k\geq K_2$. Set $K=\max(K_1,K_2)$.

Consider any indices $q\geq p>m_{K-1}$. Let $i$ be the block containing $p$, and $j$ the block containing $q$. Then $i,j\geq K$. If $i=j$, the sum from $p$ to $q$ is the difference of two partial sums measured from the start of block $i$. Each has magnitude at most $d_i$, so $$ \left|\sum_{n=p}^{q}a_n\right|\leq2d_i<\frac{\varepsilon}{2}<\varepsilon. $$ If $i<j$, write $P_i=\sum_{n=m_{i-1}+1}^{p-1}a_n$, taking $P_i=0$ when $p=m_{i-1}+1$, and $Q_j=\sum_{n=m_{j-1}+1}^{q}a_n$. Then $|P_i|\leq d_i$ and $|Q_j|\leq d_j$, and splitting the sum at the block boundaries gives $$ \sum_{n=p}^{q}a_n =\sum_{k=i}^{j-1}B_k-P_i+Q_j. $$ Consequently, $$ \left|\sum_{n=p}^{q}a_n\right| \leq\left|\sum_{k=i}^{j-1}B_k\right|+d_i+d_j <\frac{\varepsilon}{2}+\frac{\varepsilon}{4}+\frac{\varepsilon}{4} =\varepsilon. $$ In either case every sufficiently late finite block of the original series has sum of magnitude less than $\varepsilon$. The Cauchy Criterion for Series now implies that $\sum a_n$ converges. $\square$

The two conditions do different jobs. Convergence of $\sum B_k$ controls sums that begin and end at block boundaries. The requirement $d_k\to0$ controls the part of a block that may be left over when a finite tail begins or ends between boundaries. Neither condition should be omitted.

Grouping Terms and Its Limitation

One useful consequence is that grouping a convergent series into consecutive finite blocks preserves both convergence and the sum. This fact is often used to simplify expressions, but it is important to distinguish it from the converse: convergence of the grouped series alone does not generally prove convergence of the original series.

Theorem (Grouping Consecutive Terms of a Convergent Series): If $\sum_{n=1}^{\infty}a_n=S$ converges and $0=m_0<m_1<m_2<\cdots$ with $m_k\to\infty$, then $\sum_{k=1}^{\infty}B_k$ converges to $S$, where $B_k=\sum_{n=m_{k-1}+1}^{m_k}a_n$.

Proof. For every $K$, the finite block-sum identity gives $$ \sum_{k=1}^{K}B_k=\sum_{n=1}^{m_K}a_n=S_{m_K}. $$ Because $m_K\to\infty$ and $S_N\to S$, the subsequence $S_{m_K}$ tends to $S$. These are exactly the partial sums of the grouped series. Hence the grouped series converges to $S$. $\square$

Worked Example: Grouped Sums Converge but the Original Series Does Not

Take the terms $a_{2k-1}=1$ and $a_{2k}=-1$ for every $k\geq1$, and group them in pairs. Every grouped term is $$ B_k=a_{2k-1}+a_{2k}=1-1=0, $$ so the grouped series is the zero series and converges. But the original partial sums satisfy $S_{2k}=0$ and $S_{2k-1}=1$ for every $k$. They do not converge: the even partial sums are constantly $0$, while the odd partial sums are constantly $1$.

The within-block excursions are $d_k=1$, not a sequence tending to zero. In particular, starting at the first term of any pair gives a one-term block sum equal to $1$. The Cauchy Criterion therefore fails for the original series. This example shows exactly why convergence of grouped totals is not enough.

Worked Example: Pairwise Cancellation with Vanishing Excursions

For $k\geq1$, define $$ a_{2k-1}=\frac{1}{k}, \qquad a_{2k}=-\frac{1}{k}+\frac{1}{k^2}, $$ and group consecutive terms in pairs. The sum of pair $k$ is $$ B_k=a_{2k-1}+a_{2k} =\frac{1}{k}-\frac{1}{k}+\frac{1}{k^2} =\frac{1}{k^2}. $$ Thus the grouped series converges by the p-Series Convergence Criterion with $p=2$.

The only nonzero partial sum measured from the start of pair $k$ before its endpoint is $1/k$; at the endpoint the partial sum is $1/k^2$. Since $1/k^2\leq1/k$, it follows that $d_k=1/k$, and hence $d_k\to0$. The Block Criterion with Within-Block Control proves that the original series converges.

This convergence depends on cancellation within each pair. In fact, the series is not absolutely convergent: its sum of magnitudes includes the odd-indexed terms $\sum_{k=1}^{\infty}1/k$, which diverges. Thus this construction gives a conditionally convergent series, while also illustrating how a suitable block decomposition can establish convergence directly.

Using the Criterion Carefully

The Cauchy criterion is especially useful when a series has no convenient closed form for its sum, when its signs vary, or when a comparison test is not immediately available. Instead of estimating the whole sequence of partial sums against an unknown limit, one estimates differences of partial sums, equivalently finite blocks. A proof should make clear which starting indices and endpoints are being controlled and why the estimate is uniform over all such choices.

A common mistake is to show only that the terms tend to zero. The Necessary Condition for Series Convergence says that this is required for convergence, but a small individual term does not ensure that a long finite block has a small sum. The harmonic-series example demonstrates the distinction: its terms tend to zero, yet blocks with endpoints $m+1$ and $2m$ have sums at least $1/2$. Another mistake is to check only blocks aligned with a chosen grouping. The example 'Grouped Sums Converge but the Original Series Does Not' shows that aligned totals can conceal large excursions inside the blocks.

Takeaway: A series converges exactly when every sufficiently late finite block has small sum. Grouped block totals control sums at block boundaries, while vanishing within-block excursions ensure that sums beginning or ending inside a block are controlled too.

Check Your Understanding

Use the Cauchy criterion and the block results to answer the following questions.

What must be true about the starting index and endpoint quantifiers in the Cauchy Criterion for Series?
How does the block from $m+1$ through $2m$ show that the harmonic series fails the Cauchy condition?
Why does convergence of grouped block totals alone not ensure convergence of the original series?
In the Block Criterion with Within-Block Control, what does $d_k$ measure?
For the paired series with terms $1/k$ and $-1/k+1/k^2$, what are the grouped total and the within-block excursion?