The Unapologetic Mathematician

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Convergence Tests for Infinite Series

Now that we’ve seen infinite series as improper integrals, we can immediately import our convergence tests and apply them in this special case.

Take two sequences a_k and b_k with b_k\geq a_k\geq0 for all k beyond some point N. Now if the series \sum_{k=0}^\infty a_k diverges then the series \sum_{k=0}^\infty b_k does too, and if the series \sum_{k=0}^\infty b_k converges to b then the series of \sum_{k=0}^\infty a_k converges to a\leq b.

[UPDATE]: I overstated things a bit here. If the series of b_k converge, then so does that of a_k, but the inequality only holds for the tail beyond N. That is:

\displaystyle\sum\limits_{k=N}^\infty a_k\leq\sum\limits_{k=N}^\infty b_k

but the terms of the sequence a_k before N may, of course, be so large as to swamp the series of b_k.

If we have two nonnegative sequences a_k and b_k so that \lim\limits_{k\rightarrow\infty}\frac{a_k}{b_k}=c\neq0 then the series \sum_{k=0}^\infty a_k and \sum_{k=0}^\infty b_k either both converge or both diverge.

We read in Cauchy’s condition as follows: the series \sum_{k=0}^\infty a_k converges if and only if for every \epsilon>0 there is an N so that for all n\geq m\geq N the sum \left|\sum_{k=m}^n a_k\right|<\epsilon.

We also can import the notion of absolute convergence. We say that a series \sum_{k=0}^\infty a_k is absolutely convergent if the series \sum_{k=0}^\infty|a_k| is convergent (which implies that the original series converges). We say that a series is conditionally convergent if it converges, but the series of its absolute values diverges.

April 25, 2008 - Posted by | Analysis, Calculus

6 Comments »

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