# The Unapologetic Mathematician

## Associativity in Series II

I’m leaving for DC soon, and may not have internet access all day. So you get this now!

We’ve seen that associativity doesn’t hold for infinite sums the way it does for finite sums. We can always “add parentheses” to a convergent sequence, but we can’t always remove them.

The first example we mentioned last time. Consider the series with terms $a_k=(-1)^k$:

$\displaystyle\sum\limits_{k=0}^\infty a_k=1+(-1)+1+(-1)+...$

Now let’s add parentheses using the sequence $d(j)=2j+1$. Then $b_j=(-1)^{(2(j-1)+1)+1}+(-1)^{2j+1}=1+(-1)=0$. That is, we now have the sequence

$\displaystyle\sum\limits_{j=0}^\infty b_j=(1+(-1))+(1+(-1))+...=0+0+...=0$

So the resulting series does converge. However, the original series can’t converge.

The obvious fault is that the terms $a_k$ don’t get smaller. And we know that $\lim\limits_{k\rightarrow\infty}a_k$ must be zero, or else we’ll have trouble with Cauchy’s condition. With the parentheses in place the terms $b_j$ go to zero, but when we remove them this condition can fail. And it turns out there’s just one more condition we need so that we can remove parentheses.

So let’s consider the two series with terms $a_k$ and $b_j$, where the first is obtained from the second by removing parentheses using the function $d(j)$. Assume that $\lim_{k\rightarrow\infty}a_k=0$, and also that there is some $M>0$ so that each of the $b_j$ is a sum of fewer than $M$ of the $a_k$. That is, $d(j+1)-d(j). Then the series either both diverge or both converge, and if they converge they have the same sum.

We set up the sequences of partial sums

$\displaystyle s_n=\sum\limits_{k=0}^na_k$
$\displaystyle t_m=\sum\limits_{j=0}^mb_j$

We know from last time that $t_m=s_{d(m)}$, and so if the first series converges then the second one must as well. We need to show that if $t=\lim\limits_{m\rightarrow\infty}t_m$ exists, then we also have $\lim\limits_{n\rightarrow\infty}s_n=t$.

To this end, pick an $\epsilon>0$. Since the sequence of $t_m$ converge to $t$, we can choose some $N$ so that $\left|t_m-t\right|<\frac{\epsilon}{2}$ for all $m>N$. Since the sequence of terms $a_k$ converges to zero, we can increase $N$ until we also have $\left|a_k\right|<\frac{\epsilon}{2M}$ for all $k>N$.

Now take any $n>d(N)$. Then $n$ falls between $d(m)$ and $d(m+1)$ for some $m$. We can see that $m\geq N$, and that $n$ is definitely above $N$. So the partial sum $s_n$ is the sum of all the $a_k$ up through $k=d(m+1)$, minus those terms past $k=n$. That is

$\displaystyle s_n=\sum\limits_{k=0}^na_k=\sum\limits_{k=0}^{d(m+1)}a_k-\sum\limits_{n+1}^{d(m+1)}a_k$

But this first sum is just the partial sum $t_{m+1}$, while each term of the second sum is bounded in size by our assumptions above. We check

$\displaystyle\left|s_n-t\right|=\left|(t_{m+1}-t)-\sum\limits_{n+1}^{d(m+1)}a_k\right|\leq\left|t_{m+1}-t\right|+\sum\limits_{n+1}^{d(m+1)}\left|a_k\right|$

But since $n$ is between $d(m)$ and $d(m+1)$, there must be fewer than $M$ terms in this last sum, all of which are bounded by $\frac{\epsilon}{2M}$. So we see

$\displaystyle\left|s_n-t\right|<\frac{\epsilon}{2}+M\frac{\epsilon}{2M}=\epsilon$

and thus we have established the limit.

May 7, 2008 Posted by | Analysis, Calculus | 4 Comments