Divergent Series Finite Sum

Question

Something similar is probably asked somewhere else already.

How are we able to reduce some divergent series to finite sum?

As an example, 1 + r + $r^2$ + $r^3$ + ... = $\frac{1}{1-r}$. This can be proven by manipulating the series as follows:-

Let S = 1 + r + $r^2$ + $r^3$ + ...

Then $r*S$ = r + $r^2$ + $r^3$ + ...

Thus $S - r*S$ = 1 $=>$ S = $\frac{1}{1-r}$

So, when I put r = 2, I get the non-nonsensical answer that 1 + 2 + 4 + ... = -1. The sum is only correct when |r| < 1.

Why do I get this answer? Did I make some incorrect assumptions while doing this exercise?

The same technique is also used to find the nth Fibonacci number formula. We takes the Fibonacci numbers as coefficient of an infinite series, do a similar kind of manipulation and end up with a closed form of the series which we then use to find the nth term formula.

Let F = 1 + x + 2x + $3x^2$ + $5x^3$ + $8x^4$ + ...

Then using similar reordering of terms as above,

F = 1 + $F*x$ + $F*x^2$ $=>$ F = $\frac{1}{1-x-x^2}$

The above is used to find the nth Fibonacci term. Obviously if you plug in some value of x, you will get some nonsensical answer. But we still use this Math to prove some results. Can anyone shed some light on this?

It seems, something is wrong and right at the same time. Help me understand this.

Answer 1

You have an error in the partial sums. Let

$S = 1 + r + r^2 + \cdots + r^N$ ,

then $S - r S = 1 - r^{N+1}$ and

$$ S = \frac{1 - r^{N+1}}{1-r}$$

Now it should become clear that the limit $N \to \infty$ is only defined for $|r|<1$.

Answer 2

$ A_n=\sum_{i=1}^nx^n=\frac{1-x^{n+1}}{1-x}$

Now if $|x|<1$ then $lim_{n \longrightarrow \infty}A_n=\frac{1}{x-1}$

If $|x| \geqslant 1$ then the sum diverges.