# The Unapologetic Mathematician

## 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.

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Posted by John Armstrong | Analysis, Calculus

## 6 Comments	»

1. [...] on and the fact that is decreasing imply that , and the series clearly converges. Thus by the comparison test, the series converges absolutely, and our result follows. This is called Dirichlet’s test [...]

Pingback by | May 1, 2008 | Reply

2. [...] if we set , this tells us that . Then the comparison test with the geometric series tells us that [...]

Pingback by | May 5, 2008 | Reply

3. [...] don’t get smaller. And we know that must be zero, or else we’ll have trouble with Cauchy’s condition. With the parentheses in place the terms go to zero, but when we remove them this condition can [...]

Pingback by | May 7, 2008 | Reply

4. [...] if is only conditionally convergent, I say that we can rearrange the series to give any value we want! In fact, given (where these [...]

Pingback by | May 8, 2008 | Reply

5. [...] of Complex Series Today, I want to note that all of our work on convergence of infinite series carries over — with slight modifications — to complex [...]

Pingback by | August 28, 2008 | Reply

6. [...] the series of the converges, Cauchy’s condition for series of numbers tells us that for every there is some so that when and are bigger than , . But now when we [...]

Pingback by | September 9, 2008 | Reply

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