2.1 Sequence limits and proof techniques

Quantifiers in convergence

Definition

Convergence of a real sequence

A sequence $(x_n)$ converges to $L\in\mathbb R$ if for every $\varepsilon>0$ there exists $N\in\mathbb N$ such that for every $n\ge N$, $|x_n-L|<\varepsilon$. Write $x_n\to L$. The index $N$ may depend on $\varepsilon$ but must work for all later indices.

To negate convergence to $L$, there must be some $\varepsilon_0>0$ such that for every $N$ there is an $n\ge N$ with $|x_n-L|\ge\varepsilon_0$. A few early terms do not affect convergence; a recurring failure arbitrarily late does.

Worked example

Construct an epsilon–N proof

For $x_n=(3n+1)/(n+2)$, guess $L=3$. Since

$$\left|\frac{3n+1}{n+2}-3\right|=\frac5{n+2}<\frac5n,$$

choose $N>5/\varepsilon$. Then every $n\ge N$ gives $|x_n-3|<\varepsilon$. The guess is motivation; the inequality and quantified choice establish the proof.

Theorem

Uniqueness, boundedness and limit laws

A convergent real sequence has a unique limit and is bounded. If $x_n\to x$ and $y_n\to y$, then $x_n+y_n\to x+y$, $x_ny_n\to xy$, and $x_n/y_n\to x/y$ when $y\ne0$ and the quotient is defined eventually. If $x_n\le z_n\le y_n$ eventually and both outer sequences converge to $L$, then $z_n\to L$.

Proof

Uniqueness and boundedness

If $x_n\to L$ and $x_n\to M$, then for any $\varepsilon>0$, sufficiently late terms satisfy both $|x_n-L|<\varepsilon/2$ and $|x_n-M|<\varepsilon/2$. The triangle inequality yields $|L-M|<\varepsilon$. Since this holds for every positive $\varepsilon$, $L=M$. For boundedness choose $N$ so that $|x_n-L|<1$ for $n\ge N$. The finite initial terms and the bound $|x_n|\le|L|+1$ for the tail together give a finite bound for the whole sequence.

The converse of boundedness is false: $(-1)^n$ is bounded but does not converge. If it converged to $L$, its even subsequence would converge to $1$ and its odd subsequence to $-1$, violating uniqueness.

Proof

Product law mechanism

A convergent sequence $(y_n)$ is bounded, say $|y_n|\le M$. Write

$$|x_ny_n-xy|\le|y_n||x_n-x|+|x||y_n-y|.$$

Both terms tend to zero. For the quotient, $y_n\to y\ne0$ implies $|y_n|\ge|y|/2$ eventually; this keeps the denominator away from zero and gives $1/y_n\to1/y$.

Monotone sequences

Theorem

Monotone convergence theorem

Every increasing real sequence bounded above converges to the supremum of its set of terms. Every decreasing sequence bounded below converges to the infimum.

Proof

From supremum to eventual closeness

Let $s=\sup\{x_n:n\in\mathbb N\}$. Given $\varepsilon>0$, the supremum criterion gives an index $N$ with $s-\varepsilon<x_N\le s$. Monotonicity gives $s-\varepsilon<x_N\le x_n\le s$ for all $n\ge N$, proving $x_n\to s$.

Worked example

A recursive sequence

Let $x_1=1$ and $x_{n+1}=(x_n+2)/2$. Induction gives $x_n<2$, and $x_{n+1}-x_n=(2-x_n)/2>0$, so the sequence increases and is bounded above. It converges to some $L$. Taking limits in the recurrence yields $L=(L+2)/2$, hence $L=2$. Solving the fixed-point equation without first proving convergence would not by itself show the sequence has a limit.

Method comparison

Use epsilon–N estimates for a direct proof or a prescribed error. Use limit laws once component limits are established. Use squeeze when an absolute bound is easier than the exact expression. For a recurrence, seek monotonicity and a bound, prove convergence, then solve the fixed-point equation. To disprove convergence, find two subsequences with distinct limits or establish the logical negation.

Practice

1. Prove $1/n\to0$ directly.
2. Prove $(2n+3)/(n+1)\to2$.
3. Show $(-1)^n$ does not converge.
4. Let $a_1=3$ and $a_{n+1}=(a_n+1)/2$. Prove convergence and find the limit.

Full solutions

1. Given $\varepsilon>0$, choose $N>1/\varepsilon$. For every $n\ge N$, $|1/n|\le1/N<\varepsilon$.
2. The difference from $2$ is $1/(n+1)$. Choose $N>1/\varepsilon$; for $n\ge N$ the difference is below $\varepsilon$.
3. The even subsequence is constantly $1$ and the odd subsequence constantly $-1$. If the whole sequence converged, both subsequences would have the same limit, impossible.
4. By induction $a_n>1$, and $a_{n+1}-a_n=(1-a_n)/2<0$. It decreases and is bounded below by $1$, so converges to $L$. Limit laws give $L=(L+1)/2$, hence $L=1$.

Synthesis

Convergence is an eventual, quantified assertion. Completeness turns monotonicity plus a bound into existence of a limit; estimates and limit laws then identify it. Boundedness alone does not force convergence.