So through as I've been venturing through baby Rudin I came upon his definition of a cauchy sequence:

A sequence $\{ p_n \}$ in some metric space $X$ is said to be cauchy if $$ \forall \; \epsilon > 0 \; \exists \; N \in \mathbb{N} \; s.t. d(p_n, p_m) < \epsilon \; \forall \; n,m \ge N $$ He then talks about the cauchy criterion for convergence being that a cauchy sequence converges to a point in its contained metric space (similarly this metric space would be called complete). He goes on to say that the cauchy criterion for convergence for series can be restated as the following:

A series $\sum a_n$ converges if and only if $$ \forall \; \epsilon > 0 \; \exists \; N \in \mathbb{N} \; s.t. \left\lvert \sum_{k=n}^{m} a_k \right\rvert \le \epsilon \; \forall \; n,m \ge N $$ Note that these series live in $\mathbb{R}^k$ I can see how one would get

A series $\sum a_n$ converges if and only if $$ \forall \; \epsilon > 0 \; \exists \; N \in \mathbb{N} \; s.t. \left\lvert \sum_{k=n}^{m} a_k \right\rvert < \epsilon \; \forall \; n,m \ge N $$

but I don't understand why the new definition has a $\le$ rather than a $<$

Could anybody advise me as to why?

Thanks in advanced!

  • 3
    $\begingroup$ It does not matter; passing to $\epsilon/2$ will allow you to prove the second formulation from the first. $\endgroup$
    – Lord_Farin
    May 31, 2013 at 9:23
  • $\begingroup$ @Lord_Farin put it in an answer and you get it - cant believe I overthought that so much $\endgroup$
    – DanZimm
    May 31, 2013 at 10:16

1 Answer 1


The two formulations are equivalent; proving the first from the second is trivial.

For the other, apply the statement of 1 to $\epsilon/2$; then the $N$ obtained will suffice for $\epsilon$ in formulation 2.

Occasionally, allowing equality will simplify one's proof (because it makes the choice of $N = N(\epsilon)$ easier).

In fact, we can do better than that, and give a useful form for determining convergence that allows you to skip right through tedious computations when the sufficiency of $N$ is apparent -- typically of the form "when $N$ satisfies this complicated inequality involving $N$ and $\epsilon$".

For functions $f,g$, define a variant of Big O notation:

$f \in \mathcal O(g; 0)$ iff there exist $\delta$ and $C$ such that $|x|<\delta$ implies $|f(x)| < C\, |g(x)|$.

Now we have the following result (where $\epsilon$ in $\mathcal O(\epsilon;0)$ signifies the identity function):

Let $(p_k)_k$ be a real sequence (or, for that matter, a sequence in any metric space). Then $(p_k)_k$ is Cauchy iff there exists a function $f(\epsilon) \in \mathcal O(\epsilon; 0)$ such that:

$$\forall \epsilon > 0: \exists N: \forall m,n>N: d(p_m,p_n) < f(\epsilon)$$

The proof is not hard; expanding the given property of $f$ is really all there is to it. Of course, we can apply this to the sequence of partial sums of a series to obtain the Cauchy criterion for convergence of series (in a suitable space, e.g. a complete normed vector space, or the more down to earth $\Bbb R^k$).


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