Advanced Sequence Proof Mastery

Proofs About Sequences Need More Than the Right Idea

A sequence proof can have the correct intuition and still fail because its indices or quantifiers are not handled precisely. In an epsilon-\(N\) argument, the index chosen must work for every later term. In a subsequence argument, information on selected indices must be connected to the terms that were not selected. The previous tutorial studied rates of convergence; here we use that same attention to estimates to develop two proof techniques for establishing convergence.

The first technique controls a sequence by comparison with another sequence whose limit is known. The second divides the indices into blocks and controls every term in a block by its initial term. Both techniques make explicit what an informal phrase such as “the error is small” must mean: which error, how small, and for which indices?

Splitting the Error Budget

Suppose a target sequence is close to a simpler sequence, and the simpler sequence is close to its limit. The triangle inequality adds those two errors. To make their sum less than a prescribed \(\varepsilon\), it is enough to make each error less than \(\varepsilon/2\). This elementary choice is a useful proof technique because it turns two separate estimates into one convergence conclusion.

Theorem (Vanishing Perturbation Principle): Suppose \(b_n\to L\in\mathbb{R}\) and \(a_n-b_n\to0\). Then \(a_n\to L\).

Proof. Let \(\varepsilon>0\). Since \(b_n\to L\), there is an index \(N_1\) such that

|b_n-L|<\frac{\varepsilon}{2} \qquad\text{for every }n\geq N_1.

Since \(a_n-b_n\to0\), there is an index \(N_2\) such that

|a_n-b_n|<\frac{\varepsilon}{2} \qquad\text{for every }n\geq N_2.

Set \(N=\max\{N_1,N_2\}\). For every \(n\geq N\), the triangle inequality gives

|a_n-L| \leq |a_n-b_n|+|b_n-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

This is the definition of \(a_n\to L\), proving the result. \(\square\)

Worked Example: A Small Oscillation Does Not Change the Limit

Define \(a_n=\frac{n}{n+1}+\frac{(-1)^n}{n+1}\) for \(n\in\mathbb{N}_0\), and set \(b_n=\frac{n}{n+1}\). Since

|b_n-1| =\left|\frac{n}{n+1}-1\right| =\frac{1}{n+1}\longrightarrow0,

we have \(b_n\to1\). The difference between the sequences satisfies

|a_n-b_n| =\left|\frac{(-1)^n}{n+1}\right| =\frac{1}{n+1}\longrightarrow0,

because \(|(-1)^n|=1\) for every \(n\). The Vanishing Perturbation Principle now gives \(a_n\to1\). The argument does not require deciding separately how the even and odd terms behave; it bounds the oscillation uniformly by an error tending to zero.

The two estimates in the proof need not have the same natural size. The choice \(\varepsilon/2\) is convenient, not mandatory: any positive tolerances whose sum is at most \(\varepsilon\) will work. What matters is choosing the final index large enough for both estimates at once. Taking the maximum of finitely many indices is a standard way to ensure this.

Convergence from Control Within Blocks

Sometimes a sequence is easiest to understand at selected indices, called block anchors. Knowing that the anchor terms converge is not, by itself, enough to control the rest of the sequence. But if every term in a block is uniformly close to that block’s anchor, and these within-block errors tend to zero, then the selected terms do determine the limit of the whole sequence.

Theorem (Block-Anchor Convergence Criterion): Let \(n_0<n_1<n_2<\cdots\) be nonnegative integers, and let \(L\in\mathbb{R}\). Suppose \(a_{n_k}\to L\). For each \(k\), define \(\delta_k=\max\{|a_n-a_{n_k}|:n_k\leq n<n_{k+1}\}\). If \(\delta_k\to0\), then \(a_n\to L\).

Proof. Let \(\varepsilon>0\). Since \(a_{n_k}\to L\), there is an index \(K_1\) such that

|a_{n_k}-L|<\frac{\varepsilon}{2} \qquad\text{for every }k\geq K_1.

Since \(\delta_k\to0\), there is an index \(K_2\) such that \(\delta_k<\varepsilon/2\) for every \(k\geq K_2\). Set \(K=\max\{K_1,K_2\}\). The strictly increasing integer indices \(n_k\) tend to infinity. Consequently, every integer \(n\geq n_K\) lies in one of the blocks \([n_k,n_{k+1})\) for some \(k\geq K\). For that \(k\), the definition of \(\delta_k\) gives

|a_n-a_{n_k}|\leq\delta_k<\frac{\varepsilon}{2}.

The triangle inequality therefore yields

|a_n-L| \leq |a_n-a_{n_k}|+|a_{n_k}-L| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2} =\varepsilon.

This holds for every \(n\geq n_K\), so \(a_n\to L\). \(\square\)

Worked Example: Convergence from Square-Indexed Anchors

For each integer \(k\geq1\), consider the block of indices \(k^2\leq n<(k+1)^2\). Define

a_n=\frac{k}{k+1}+\frac{n-k^2}{k^2(k+1)} \qquad\text{when }k^2\leq n<(k+1)^2,

and set \(a_0=0\). The blocks cover all positive integer indices: the first starts at \(1\), each block ends just before the next starts, and \(k^2\to\infty\). The anchor at \(n_k=k^2\) is

a_{k^2}=\frac{k}{k+1}\longrightarrow1.

Within the \(k\)-th block, \(0\leq n-k^2<(k+1)^2-k^2=2k+1\). Since \(n-k^2\) is an integer, \(n-k^2\leq2k\). Thus every term in the block satisfies

|a_n-a_{k^2}| =\frac{n-k^2}{k^2(k+1)} \leq\frac{2k}{k^2(k+1)} =\frac{2}{k(k+1)} \longrightarrow0.

The block-anchor criterion applies with \(L=1\), proving \(a_n\to1\). The proof controls all terms in each block, not just the square-indexed terms.

When a Useful-Looking Estimate Is Not Enough

A common mistake is to assume that small changes between consecutive terms force convergence. The condition \(a_{n+1}-a_n\to0\) only says that adjacent terms eventually differ by little. It does not prevent small changes from accumulating over many indices.

Worked Example: Small Successive Differences Do Not Guarantee Convergence

Let \(a_n=\sum_{j=1}^{n}\frac{1}{j}\) for \(n\geq1\), with \(a_0=0\). Its successive differences satisfy

a_{n+1}-a_n=\frac{1}{n+1}\longrightarrow0.

Nevertheless, the sequence is unbounded. For each integer \(r\geq1\), consider the terms with \(2^{r-1}<j\leq2^r\). There are \(2^{r-1}\) such terms, and each is at least \(1/2^r\). Hence

\sum_{j=2^{r-1}+1}^{2^r}\frac{1}{j} \geq 2^{r-1}\frac{1}{2^r} =\frac12.

The first term is \(1\), so adding the contributions from the first \(m\) such groups gives \(a_{2^m}\geq1+m/2\). This tends to \(+\infty\), and therefore the full sequence is unbounded and cannot converge to a finite real number. Small successive differences alone do not give the uniform control across a long block that the block-anchor criterion requires.

A related error is to find one convergent subsequence and conclude that the original sequence converges. For example, \(a_n=(-1)^n\) has the subsequence \(a_{2k}=1\), which converges to \(1\), but the full sequence does not converge: its even terms equal \(1\) and its odd terms equal \(-1\). A subsequence supplies information only at its selected indices. To prove convergence of the whole sequence, one needs additional control over the omitted terms, such as the block estimates above.

A Practical Proof Check

Before finishing a sequence proof, check whether each estimate covers every index the conclusion requires. In an epsilon-\(N\) proof, confirm that the final \(N\) works for every later \(n\). In a block argument, confirm that the blocks cover the entire tail and that the error bound is uniform across each block. In a perturbation argument, identify both errors and verify that both are small beyond one common index.

These checks distinguish a complete proof from a plausible sketch. Convergence is a statement about all sufficiently late terms, not merely a selected subsequence, a typical term, or the size of one-step changes. When the proof identifies exactly how every late term is controlled, the conclusion follows directly from the definition.

Check Your Understanding

Use the proof techniques and examples in this tutorial to answer the following questions.

In the Vanishing Perturbation Principle, why is it enough to make each of the two errors smaller than \(\varepsilon/2\)?
In the block-anchor criterion, what two quantities must become small, and how are they combined to control an arbitrary term?
For the square-indexed example, why does every integer \(n\geq1\) belong to one of the stated blocks?
Why does a convergent subsequence of a sequence not, by itself, establish convergence of the full sequence?
What feature of the harmonic partial sums shows that small successive differences do not ensure convergence?