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The sequence of functions $(f_n)$ converges uniformly to $f$ on $E$ ifThe same $N$ must work for every $x$. Equivalently, the supremum norm of $f_n-f$ tends to zero. This is stronger than pointwise convergence, where $N$ may depend on $x$.

A sequence is an ordered family $a_1,a_2,\dots$, formally a function from the positive integers into a chosen set. The indices record position and allow repeated values. A series formed from that sequence studies the partial sumsWriting ${\sum }_{n=1}^{\infty }{a}_n=S$ means that the sequence $(S_N)$ has finite limit $S$. Thus convergence of terms and convergence of their cumulative sums are different assertions.

For $a_n=2^{-n}$, the terms tend to zero and the geometric series has $S_N=1-2^{-N}\to 1$. For $a_n=1/n$, the terms also tend to zero, but the harmonic series diverges. A necessary condition for series convergence follows from $a_n=S_n-S_{n-1}$; it is not sufficient. The Cauchy criterion for series instead controls every sufficiently late finite tail.

For a sequence of functions, there are two further choices: pointwise convergence checks a separate limit at each input, while uniform convergence requires a common error bound over the domain. These distinctions determine when infinite sums may be integrated or differentiated.

A sequence of functions $f_n:E\to \mathbb{R}$ has a uniform limit if and only if, for every $\epsilon >0$, there is $N$ such that $|f_n(x)-f_m(x)|<\epsilon$ for every $m,n\ge N$ and every $x\in E$. Necessity follows from the triangle inequality. For sufficiency, completeness of the real numbers first gives a pointwise limit $f$. Letting $m\to \infty$ in the bound with tolerance $\epsilon /2$ gives $|f_n(x)-f(x)|\le \epsilon /2$ uniformly in $x$. The same proof works for a complete metric space as codomain; completeness is essential.

A function $f:E\to \mathbb{R}$ is the uniform limit of a sequence of functions $(f_n)$ whenEquivalently the supremum error ${\sup }_{x\in E}|f_n(x)-f(x)|$ tends to zero. The same index works for all inputs, including inputs chosen differently for each $n$. A uniform limit of continuous functions is continuous, by approximating both endpoint values with one common approximant.

For $f_n(x)=x^n$ on $[0,a]$, $0<a<1$, the zero function is a uniform limit because the supremum error is $a^n$. On $[0,1]$, the pointwise limit instead equals zero below one and one at one. Its supremum error remains one for every $n$, so there is no uniform limit there. The open supremum below the endpoint need not be attained to obstruct uniform convergence.

For each positive integer $n$, put $f_n(t)=\sin (2\pi nt)/n$ on $[0,1]$. Its supremum norm is exactly $1/n$, so this sequence of functions converges to zero by uniform convergence. Nevertheless,Each of its $n$ complete sine oscillations contributes $4/n$: rise from zero to $1/n$, fall through zero to $-1/n$, then return to zero. Taking a partition of an interval through all extrema attains $4$. Every further partition refinement merely subdivides monotone function segments, so the same reasoning as in partition example of total variation proves the upper bound too.

For a continuously differentiable function on a closed bounded interval, total variation also satisfiesApplied here, the derivative is $2\pi \cos (2\pi nt)$, and integrating its absolute value over $n$ periods gives $4$. The differentiability hypothesis matters: the partition definition applies much more widely than this derivative formula.^[ref] Small pointwise size and small accumulated movement are different approximation goals.