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One particular series that shows
up in many applications,

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examples, or problems is
the geometric series.

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[In] the geometric series, we
are given a certain number,

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alpha, and we want to sum all
the powers of alpha, starting

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from the 0th power, which is
equal to 1, the first power,

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and so on, and this gives
us an infinite series.

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It's the sum of alpha
to the i where i

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ranges from 0 to infinity.

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Now, for this series to
converge, we need subsequent

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terms, the different terms in
the series, to become smaller

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and smaller.

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And for this reason, we're going
to make the assumption

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that the number alpha is less
than 1 in magnitude, which

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implies that consecutive
terms go to zero.

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Let us introduce
some notation.

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Let us denote the infinite sum
by s, and we're going to use

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that notation shortly.

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One way of evaluating this
series is to start from an

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algebraic identity, namely
the following.

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Let us take 1 minus alpha and
multiply it by the terms in

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the series, but going only up
to the term alpha to the n.

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So it's a finite series.

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We do this multiplication, we
get a bunch of terms, we do

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the cancellations, and what is
left at the end is 1 minus

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alpha to the power n plus 1.

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What we do next is we take the
limit as n goes to infinity.

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On the left hand side, we have
the term 1 minus alpha, and

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then the limit of this finite
series is by definition the

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infinite series, which
we're denoting by s.

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On the right hand side,
we have the term 1.

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How about this term?

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Since alpha is less than 1 in
magnitude, this converges to 0

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as alpha goes to infinity,
so that term disappears.

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We can now solve this relation,
and we obtain that s

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is equal to 1 over 1 minus
alpha, and this is the formula

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for the infinite geometric
series.

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There's another way of deriving
the same result,

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which is interesting, so let
us go through it as well.

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The infinite geometric series
has one first term and then

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the remaining terms, which is
a sum for i going from 1 to

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infinity of alpha to the i.

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Now, we can take a factor of
alpha out of this infinite sum

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and write it as 1 plus alpha,
the sum of alpha to the i, but

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because we took out one factor
of alpha, here, we're going to

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have smaller powers.

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So now the sum starts from 0
and goes up to infinity.

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Now, this is just 1 plus alpha
times s because here, we have

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the infinite geometric series.

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Therefore, if we subtract alpha
s from both sides of

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this equality, we get s times
1 minus alpha equal to 1.

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And now by moving 1 minus alpha
to the denominator, we

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get again the same expression.

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So this is an alternative way
of deriving the same result.

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However, there's one
word of caution.

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In this step, we subtracted
alpha s from both

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sides of the equation.

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And in order to do that, this
is only possible if we take

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for granted that s is
a finite number.

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So this is taken for granted
in order to carry out this

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derivation.

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This is to be contrasted with
the first derivation, in which

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we didn't have to make
any such assumption.

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So strictly speaking, for this
derivation here to be correct,

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we need to have some independent
way of verifying

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that s is less than infinity.

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But other than that, it's an
interesting algebraic trick.