What does the false infinite sum of a series mean?

Question

For any geometric series with |$r$| < 1 , I know that

$$\sum_{k=1}^{∞} ar^{k-1} =\frac{a}{1-r}$$

But if |$r$| > 1 and you try to use the formula, you'll get a weird answer. For instance:

$$4+8+16+32+64+128+... =\sum_{k=1}^{∞} (4)2^{k-1}= \frac{4}{1-2} = -4$$

That answer obviously doesn't make sense; the series diverges. So what does -4 mean? Where did it come from, and how is it related to the series? It must be significant somehow.

Answer 1

Let's look at the formula for a finite geometric series first. If $a$ is the first term, $r$ is the common ratio, and $n$ is the number of terms, then your sum is equal to $a\frac{1-r^n}{1-r}$. Now, if $|r|<1$, we can see that as $n$ goes to infinity the $r^n$ term disappears, leaving the familiar formula of $a\frac{1}{1-r}$. So, the infinite geometric series sum formula makes the assumption that $|r|<1$.

With the finite geometric series sum formula, we can rewrite as follows:

$$a\frac{1-r^n}{1-r}$$

$$=a\frac{r^n-1}{r-1}$$

$$=\frac{a}{r-1}r^n - \frac{a}{r-1}$$

Remember, an infinite sum is just the limit as $n\to\infty$ of a finite sum. Now, we know that the infinite sum assumes that $r^n$ goes to 0. If we assume that the first term goes to 0 even though it doesn't, we end up with the second term of that last line: $$\frac{-a}{r-1}$$

If you look at the terms individually, you can see that as $n\to\infty$ the first term goes to infinity. In a sense, you can think of that number you get as "ignoring infinity's contribution" to the sum (though this idea is very informal). It's what the sum would have been if the first term disappeared like it does when $|r|\lt1$.

Answer 2

This is a way of assigning meaning to an otherwise divergent sum.

Look up "divergent series."

The subject is large.

Answer 3

For a truncated series, we have

$$\sum_{k=1}^Kar^{k-1}=a\frac{1-r^K}{1-r} \tag 1$$

For $a=4$ and $r=2$, use of $(1)$ reveals

$$\sum_{k=1}^K4\left(2\right)^{k-1}=4\frac{1-2^K}{1-2}=2^{K+2}-4$$

The $-4$ appears as the "convergent component" of the divergent series (as $K\to \infty$).

Answer 4

This is an estimation! If you will look how this estimation is derived you will see immediately that the sum goes to infinity because of $r^n$ in the sum