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# Uniform Convergence of a Series Explained in Plain Words

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Why does it suffice to say that |S-S_n|< epsilon implies that S=sum(n=1 to infinity) a_n is uniformly convergent? I am looking for the reasoning behind it or some reading material on it.

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## SOLUTION This solution is FREE courtesy of BrainMass!

A series converges means that the sequence of its partial sums converges.
You can define a sequence s_n = sum_{k=1}^n a_n and if it has a limit, s, then the series converges to this limit, and the sum of the series is s.
The definition of convergence of a sequence is that for every epsilon there is an N such that for all n>N we have |s_n - s| < epsilon.

Now, if the general term of the sequence or a series contains a variable, say x, then the question of uniform and pointwise convergence come into play. We can either fix the value of x and consider the convergence of the series or the sequence at this value of x. It might happen that for all x the series converges pointwise, that is

for every x for every epsilon there is an N(x), _possibly dependent on x_ such that for all n>N we have |s_n(x)-s(x)|<epsilon.

This is pointwise convergence.

Then, for uniform convergence we require that N be independent of x, that is, we require that

for every epsilon > 0 there is an N>0 such that for every x and for every n>N we have |s_n(x)-s(x)|<epsilon.

One N works for all x -- that is the major difference between pointwise and uniform convergence.

The uniform convergence is dependent on the interval for x. Usually, a series (or a sequence) is more likely to converge uniformly on a bounded interval than to converge uniformly on the entire real line. On an interval we can bound the values of x, and can hope to estimate the partial sums by numbers independent on x. Then, if the sequence of these numbers converges, then by the M-test the series converges uniformly (because the dependence on x is removed).

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