TITLE: Assigning values to divergent series
QUESTION [22 upvotes]: I have been looking at divergent series on wikipedia and other sources and it seems people give finite "values" to specific ones. I understand that these values sometimes reflect the algebraic properties of the series in question, but do not actually represent what the series converges to, which is infinity. Why is it usefull to assign values to divergent series?
The only theory I could come up with, is this:
Say you have 2 divergent series, series' A and B, and you assign each a value, 
Series ($A=
\sum_{n=0}^\infty a_n$), which I assigned the value Q
and series ($B=
\sum_{n=0}^\infty b_n$ ), which I assigned the value P
But it just so happens that series $C=A-B=
\sum_{n=0}^\infty (a_n-b_n)$ converges.
Could that imply that the actual value of series $C$ is the difference of the two assigned values to $A$ and $B$, that is $\sum_{n=0}^\infty (a_n-b_n)=Q-P$ ?
If so, then that would make some sense to me, as to why people sometimes assign values to divergent series.

REPLY [7 votes]: There exist systems in which "divergent" series "coverge". They are called the systems of $p$-adic numbers, and for example in $3$-adic system you have
$$\sum_{k\geq0} 2\cdot 3^k=-1,$$
therefore
$$-1=\dotsm 2222222\bullet_3.$$
To show that the identity above makes some sense, let's try to multiply $(-1)\cdot(-1)$ by the "basic shool" algorithm, in the base $3$. We get
$$\begin{array}
{}&{}&\dotsm&2&2&2&2&2&2&2\\
{}&\times&\dotsm&2&2&2&2&2&2&2\\\hline
{}&{}&\dotsm&2&2&2&2&2&2&1\\
{}&+&\dotsm&2&2&2&2&2&1\\
{}&+&\dotsm&2&2&2&2&1\\
{}&+&\dotsm&2&2&2&1\\
{}&+&\dotsm&2&2&1\\
{}&+&\dotsm&2&1\\
{}&+&\dotsm&1\\
{}&+&\vdots&\vdots\\
\hline
{}&=&\dotsm&0&0&0&0&0&0&1\\
\text{where the carry was:}&{}&\dotsm&6&5&4&3&2&1&0\\
\end{array}$$