Proof of the Cauchy Criterion

From Finite Blocks to Partial Sums

The Cauchy Criterion for Series turns convergence into a condition on finite blocks of terms. In the previous tutorial, this condition was used to test convergence and to analyze grouped series. Here we prove the criterion itself. The proof connects two ideas: a finite block sum is a difference of partial sums, and every Cauchy sequence of real numbers converges.

Let $S_0=0$ and $S_N=\sum_{n=1}^{N}a_n$ for $N\geq1$. By the Block-Sum Identity, whenever $q\geq p\geq1$,

\sum_{n=p}^{q}a_n=S_q-S_{p-1}.

Thus, controlling all sufficiently late finite block sums is equivalent to controlling differences between sufficiently late partial sums. The index shift from $p$ to $p-1$ matters: a block starting at $p$ compares $S_q$ with $S_{p-1}$, not with $S_p$.

Theorem (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 $N\geq1$ such that $$ \left|\sum_{n=p}^{q}a_n\right|<\varepsilon $$ whenever $q\geq p\geq N$.

We will prove this by first making explicit the completeness fact needed in the reverse direction. A Cauchy sequence has terms that eventually lie arbitrarily close to one another. To conclude that they converge to a real number, we use the completeness of $\mathbb{R}$.

Why Cauchy Sequences of Real Numbers Converge

Theorem (Cauchy Criterion for Sequences in $\mathbb{R}$): Every Cauchy sequence of real numbers converges.

Proof. Let $(x_n)$ be a Cauchy sequence. First, it is bounded. Choose an index $N_0$ such that $|x_n-x_{N_0}|<1$ whenever $n\geq N_0$. Then $|x_n|\leq |x_{N_0}|+1$ for $n\geq N_0$. The finitely many earlier terms are also bounded, so the whole sequence is bounded.

For each $n$, define the supremum and infimum of the tail:

u_n=\sup\{x_k:k\geq n\}, \qquad \ell_n=\inf\{x_k:k\geq n\}.

These are finite real numbers because the sequence is bounded and each tail is nonempty. Since the tails shrink as $n$ increases, $(u_n)$ is nonincreasing and $(\ell_n)$ is nondecreasing. By completeness, the nonempty set $\{u_n:n\geq1\}$, which is bounded below, has an infimum. Write

L=\inf_{n\geq1}u_n.

Fix $\varepsilon>0$. Since $(x_n)$ is Cauchy, there is an $N$ such that $|x_j-x_k|<\varepsilon/2$ for all $j,k\geq N$. This implies $u_N-\ell_N\leq\varepsilon/2$: all values in the $N$-th tail are within $\varepsilon/2$ of one another, so the supremum and infimum of that tail differ by at most $\varepsilon/2$. Also $\ell_N\leq L\leq u_N$. Indeed, $u_N$ is one of the numbers whose infimum defines $L$, and every $u_n$ is at least $\ell_N$: for $n\geq N$, its tail lies above $\ell_N$, while for $n<N$, $u_n\geq u_N\geq\ell_N$. Therefore, for every $n\geq N$, both $x_n$ and $L$ lie in $[\ell_N,u_N]$, and

|x_n-L|\leq u_N-\ell_N\leq\frac{\varepsilon}{2}<\varepsilon.

Hence $x_n\to L$, proving the theorem. $\square$

This proof identifies the role of completeness: it guarantees that the tails have a real number $L$ to approach. The Cauchy property alone gives increasingly narrow tails; completeness ensures their limiting location belongs to $\mathbb{R}$.

Proof of the Series Criterion

Proof. First suppose $\sum_{n=1}^{\infty}a_n$ converges. Then its partial sums $S_N$ converge, so they form a Cauchy sequence. Given $\varepsilon>0$, choose $K$ such that

|S_r-S_s|<\varepsilon

whenever $r,s\geq K$. Set $N=K+1$. If $q\geq p\geq N$, then $q\geq K$ and $p-1\geq K$. The Block-Sum Identity gives

\left|\sum_{n=p}^{q}a_n\right| =|S_q-S_{p-1}| <\varepsilon.

This proves the finite-block condition.

Conversely, suppose that for every $\varepsilon>0$ there is an $N$ such that every block beginning at $p\geq N$ has sum of magnitude less than $\varepsilon$. We show that the partial sums form a Cauchy sequence. Given $\varepsilon>0$, choose such an $N$. If $r,s\geq N$ and $r<s$, apply the block condition with $p=r+1$ and $q=s$. Since $r+1\geq N$,

|S_s-S_r| =\left|\sum_{n=r+1}^{s}a_n\right| <\varepsilon.

If $r=s$, the difference is zero; if $s<r$, interchange the indices and use $|S_r-S_s|=|S_s-S_r|$. Thus $(S_N)$ is Cauchy. By the Cauchy Criterion for Sequences in $\mathbb{R}$, it converges to a real number. By definition, the series converges. This proves the reverse implication and completes the proof. $\square$

Both directions use the same identity, but the index choices differ. In the forward direction, the block starts at $p$, so the partial-sum indices are $p-1$ and $q$. In the reverse direction, to compare $S_r$ and $S_s$, the block must begin at $r+1$. Keeping those shifts explicit prevents a common off-by-one error.

Worked Applications of the Criterion

Worked Example: A Geometric Series by Uniform Block Bounds

Consider $\sum_{n=1}^{\infty}4^{-n}$. For any $q\geq p\geq1$, the finite geometric-sum formula gives

\sum_{n=p}^{q}4^{-n} =4^{-p}\frac{1-4^{-(q-p+1)}}{1-1/4}.

The numerator $1-4^{-(q-p+1)}$ is positive and at most $1$, so

0\leq\sum_{n=p}^{q}4^{-n} \leq\frac{4}{3}4^{-p} \leq\frac{4}{3}4^{-N} \qquad (p\geq N).

Given $\varepsilon>0$, choose $N$ large enough that $\frac{4}{3}4^{-N}<\varepsilon$. Then every block with $q\geq p\geq N$ has sum less than $\varepsilon$. The Cauchy Criterion for Series proves convergence without requiring a proposed value for the sum.

Worked Example: A Telescoping Series by Block Estimates

Let $a_n=1/(n(n+1))$. Since

\frac{1}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1},

a finite block telescopes:

\sum_{n=p}^{q}\frac{1}{n(n+1)} =\frac{1}{p}-\frac{1}{q+1}.

For $q\geq p$, this quantity is nonnegative and at most $1/p$. Therefore, if $p\geq N$,

\left|\sum_{n=p}^{q}\frac{1}{n(n+1)}\right| \leq\frac{1}{p} \leq\frac{1}{N}.

For any $\varepsilon>0$, choose an integer $N$ with $1/N<\varepsilon$. The block condition follows, so the series converges. The estimate is uniform in the endpoint $q$, as the criterion requires.

Worked Example: Terms Tend to Zero but the Criterion Fails

Consider $a_n=1/\sqrt{n}$. The terms tend to zero, but that fact alone does not control long blocks. Given any integer $N$, choose an integer $m\geq\max\{N,2\}$ 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}{\sqrt{k}}\geq\frac{1}{\sqrt{2m}}.

Adding these $m$ inequalities gives

\sum_{k=m+1}^{2m}\frac{1}{\sqrt{k}} \geq\frac{m}{\sqrt{2m}} =\sqrt{\frac{m}{2}} \geq1.

The block begins at $m+1\geq N$, yet its sum is at least $1$. Thus the Cauchy condition fails with $\varepsilon=1$, and the series diverges. This shows why small individual terms are not enough: the criterion controls every late finite block, including blocks whose lengths grow with their starting indices.

What the Proof Tells Us

The criterion is more than a convenient test. It expresses convergence as a uniform statement about all late blocks. The starting index $N$ may depend on $\varepsilon$, but once chosen, it must work for every $p\geq N$ and every finite endpoint $q\geq p$. Checking only blocks of one fixed length, or only blocks from a particular grouping, does not establish this uniform control.

The proof also separates the algebra from the completeness step. The Block-Sum Identity translates between block sums and differences of partial sums. The forward implication uses the fact that a convergent sequence is Cauchy. The reverse implication first makes the partial sums Cauchy, then uses completeness of the real numbers to obtain their limit. Without completeness, that final conclusion need not hold in the same number system.

Takeaway: A series converges exactly when every sufficiently late finite block has small sum. The equivalence follows by translating block sums into differences of partial sums and, in the reverse direction, using completeness to turn a Cauchy sequence of partial sums into a convergent one.

Check Your Understanding

Use the proof and examples to answer the following questions.

Why does a block from $p$ to $q$ equal $S_q-S_{p-1}$, rather than $S_q-S_p$?
In the forward implication, why is it useful to choose the block-condition index as $N=K+1$ when the partial sums are Cauchy from index $K$ onward?
How does the reverse implication use a block to compare two partial sums $S_r$ and $S_s$?
Where does completeness of $\mathbb{R}$ enter the proof of the Cauchy Criterion for Series?
Why does $a_n\to0$ not suffice to prove convergence of $\sum a_n$?

What You'll Learn