Sequences · Definition

Convergent sequence

real analysissequences

Before you start you need to know

Convergence is the central idea of analysis: it makes precise what it means for an infinite list of numbers to approach a value without necessarily reaching it.

Definition

A sequence \((a_n)\) converges to a number \(L\) if, for every tolerance \(\varepsilon>0\), there is an index \(N\) from which all terms are within distance \(\varepsilon\) of \(L\).

\[\forall\varepsilon>0\ \exists N:\ n\ge N\Rightarrow|a_n-L|<\varepsilon\]

The definition is a game between two players. Someone proposes a tolerance \(\varepsilon>0\), as small as they like, and you must answer with an index \(N\) such that from then on no term strays from \(L\) by more than \(\varepsilon\). The index \(N\) depends on \(\varepsilon\): the stricter the tolerance, the further out you must go. For \(a_n=1/n\) and \(L=0\), any \(N>1/\varepsilon\) works.

Geometrically, the condition says that every interval \((L-\varepsilon,L+\varepsilon)\) contains all the terms except finitely many. The first terms can behave arbitrarily, and the sequence may even equal \(L\) many times; what matters is what happens in the tail. That is why changing finitely many terms alters neither convergence nor the limit.

To prove convergence one works backwards: start from \(|a_n-L|<\varepsilon\), solve for \(n\), and that suggests which \(N\) to choose. With \(a_n=\frac{n+1}{n}\) and \(L=1\) we get \(|a_n-1|=\frac1n<\varepsilon\) when \(n>1/\varepsilon\). If no \(L\) satisfies the definition, the sequence diverges, like \((-1)^n\) or \(a_n=n\).

A common mistake is to read the definition as “the terms get closer and closer to \(L\)”; that is not necessary. The sequence \(1,\,0,\,\tfrac12,\,0,\,\tfrac13,\dots\) converges to \(0\) without approaching monotonically. Nor is approaching enough: what counts is staying, from some index on, inside every interval around \(L\).

Think about it

Prove from the definition that \(\frac{2n+1}{n}\to2\).

We have \(\left|\frac{2n+1}{n}-2\right|=\frac1n\). Given \(\varepsilon>0\), if \(N>1/\varepsilon\) and \(n\ge N\), then \(\frac1n<\varepsilon\).

Can an unbounded sequence converge?

No: from \(N\) on all terms lie in \((L-1,L+1)\), and there are only finitely many earlier terms, so the whole sequence is bounded.

Exercises 10

Compute the following limits

  1. \(\displaystyle \lim_{n\to\infty}\frac1n\)
  2. \(\displaystyle \lim_{n\to\infty}\frac{3n+1}{n+2}\)
  3. \(\displaystyle \lim_{n\to\infty}\frac{2n^2+1}{n^2+n}\)
  4. \(\displaystyle \lim_{n\to\infty}\left(\sqrt{n+1}-\sqrt n\right)\)

Problems

  1. For \(a_n=\frac1n\) and \(\varepsilon=0.01\), what is the smallest \(N\) such that \(|a_n-0|<\varepsilon\) for all \(n\ge N\)?
  2. What happens with the sequence \(a_n=(-1)^n\)?
  3. For \(a_n=\frac{n+1}{n}\), which converges to \(1\), and \(\varepsilon=0.1\), what is the smallest \(N\) that works?
  4. If \(a_n\to L\), how many terms of the sequence fall outside the interval \((L-\varepsilon,L+\varepsilon)\), for a fixed \(\varepsilon>0\)?
  5. If \(a_n\to L\), which statement is always true?
  6. For \(a_n=\frac{n}{n+1}\), which converges to \(1\), and \(\varepsilon=\frac1{1000}\), what is the smallest \(N\) such that \(|a_n-1|<\varepsilon\) for all \(n\ge N\)?

References

  • Stephen Abbott, Understanding Analysis, 2nd edition, Springer, 2015. §2.2
  • Walter Rudin, Principles of Mathematical Analysis, 3rd edition, McGraw-Hill, 1976. Cap. 3, 3.1
  • Robert G. Bartle, Donald R. Sherbert, Introduction to Real Analysis, 4th edition, Wiley, 2011. §3.1