One particular series that shows up in many applications,
examples, or problems is the geometric series.
[In] the geometric series, we are given a certain number,
alpha, and we want to sum all the powers of alpha, starting
from the 0th power, which is equal to 1, the first power,
and so on, and this gives us an infinite series.
It's the sum of alpha to the i where i
ranges from 0 to infinity.
Now, for this series to converge, we need subsequent
terms, the different terms in the series, to become smaller
and smaller.
And for this reason, we're going to make the assumption
that the number alpha is less than 1 in magnitude, which
implies that consecutive terms go to zero.
Let us introduce some notation.
Let us denote the infinite sum by s, and we're going to use
that notation shortly.
One way of evaluating this series is to start from an
algebraic identity, namely the following.
Let us take 1 minus alpha and multiply it by the terms in
the series, but going only up to the term alpha to the n.
So it's a finite series.
We do this multiplication, we get a bunch of terms, we do
the cancellations, and what is left at the end is 1 minus
alpha to the power n plus 1.
What we do next is we take the limit as n goes to infinity.
On the left hand side, we have the term 1 minus alpha, and
then the limit of this finite series is by definition the
infinite series, which we're denoting by s.
On the right hand side, we have the term 1.
How about this term?
Since alpha is less than 1 in magnitude, this converges to 0
as alpha goes to infinity, so that term disappears.
We can now solve this relation, and we obtain that s
is equal to 1 over 1 minus alpha, and this is the formula
for the infinite geometric series.
There's another way of deriving the same result,
which is interesting, so let us go through it as well.
The infinite geometric series has one first term and then
the remaining terms, which is a sum for i going from 1 to
infinity of alpha to the i.
Now, we can take a factor of alpha out of this infinite sum
and write it as 1 plus alpha, the sum of alpha to the i, but
because we took out one factor of alpha, here, we're going to
have smaller powers.
So now the sum starts from 0 and goes up to infinity.
Now, this is just 1 plus alpha times s because here, we have
the infinite geometric series.
Therefore, if we subtract alpha s from both sides of
this equality, we get s times 1 minus alpha equal to 1.
And now by moving 1 minus alpha to the denominator, we
get again the same expression.
So this is an alternative way of deriving the same result.
However, there's one word of caution.
In this step, we subtracted alpha s from both
sides of the equation.
And in order to do that, this is only possible if we take
for granted that s is a finite number.
So this is taken for granted in order to carry out this
derivation.
This is to be contrasted with the first derivation, in which
we didn't have to make any such assumption.
So strictly speaking, for this derivation here to be correct,
we need to have some independent way of verifying
that s is less than infinity.
But other than that, it's an interesting algebraic trick.