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## Integral Calculus

### Unit 5: Lesson 16

Representing functions as power series- Integrating power series
- Differentiating power series
- Integrate & differentiate power series
- Finding function from power series by integrating
- Integrals & derivatives of functions with known power series
- Interval of convergence for derivative and integral
- Converting explicit series terms to summation notation
- Converting explicit series terms to summation notation (n ≥ 2)

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# Converting explicit series terms to summation notation (n ≥ 2)

Usually, when writing a series in sigma notation, we start summing at n=1 or n=0. Sometimes though, we would like to start at n=2 or larger values. See an example here. Created by Sal Khan.

## Video transcript

Let's say that we're told that
this sum right over here, where our index starts at 2 and we
go all the way to infinity, that this infinite
series is negative 8/5 plus 16/7 minus 32/9
plus-- and we just keep going on and on forever. And so what I want to do
is to explicitly define what a sub n is here. So right now we just
say, hey, if you take the sum of a sub n
from n equals 2 to infinity, it turns out you get
this sum right over here. But let's think about what a
sub n-- how we can actually define it in terms of n. And I encourage you to pause
the video right now and try it on your own. So the first thing
that you might realize is, well, this is the number
that we're going to get. Let me write it this way. a sub 2 is equal
to negative 8/5. a sub 3 is equal to 16/7. a sub 4 is equal
to negative 32/9. And I'm just giving the sign
to the number in the numerator. Negative 8/5 is the same
thing as negative 8 over 5. Let me make that a
little bit clearer. So I'll make that a
little bit clearer. So this is negative 8/5. Obviously, this
is positive, so I don't have to really
worry about it too much. And then here, I'm just
saying negative 32/9, so it's the same thing
as negative 32 over 9. So let's see if
we can first find a pattern in the numerator. So when we go from negative
8 to 16, what's happening? Well, we're multiplying
by negative 2. Now, to go from
16 to negative 32, we're multiplying
by negative 2 again. So you might say,
OK, well, whatever we have in the numerator must
be a power of negative 2. And, all right, if you
say, well, maybe this is negative 2 squared, well,
you know that negative 8 isn't negative 2 squared. Negative 2 squared is
equal to positive 4. Negative 8-- this
right over here. Negative 8, that is equal to
negative 2 to the third power. 16 is equal to negative
2 to the fourth power. Negative 32 is equal to
negative 2 to the fifth power. So notice, our exponent
on the negative 2 is always going to be
one more than our index. Our index is 2,
our exponent is 3. Our index is 3,
our exponent is 4. Our index is 4,
our exponent is 5. So that gives a sense that at
least the numerator is going to be-- whatever our index
is, it's going to be-- so let me write this down. So a sub n is equal
to-- well, it's going to be negative 2 to
whatever index we're at, to that index plus 1 power. So that's a reasonable way
to think about our numerator. Now let's think about
our denominators. So over-- So we go from 5,
so when n is 2, we're at 5. When n is 3, we're at 7. When n is 4, we're at 9. So notice, 5 is
2 times 2 plus 1. This right over here
is 2 times 3 plus 1. This right over here
is 2 times 4 plus 1. And you should just
kind of play around with different
patterns in your head until you say, hey,
well, look, this is increasing by 2 every time. Notice, this increases
by 2 every time. But these aren't
exactly multiples of 2. These seem to be off by one more
than the multiples of 2, which is a good sign
that this is going to be 2 times our index plus 1. So we could write this as
2 times our index plus 1. And we're done. That's what a sub n is. And if we wanted to write
this series in sigma notation, we would write this
as the sum from n equals 2 to infinity
of negative 2 to the n plus 1
power over 2n plus 1. And that would equal this
series right over here.