# Converting explicit series terms to summation notation

## Video transcript

So I have the series-- negative
5/3 plus 25 over 6 minus 125 over 9 plus-- and it just keeps
going on and on and on forever. So this right over
here is an infinite sum or an infinite series, and
what I want you to do right now is to pause this video
and try to express this infinite series
using sigma notation. So I'm assuming you've given
a go at it, so let's just look at each term of the series
and let's see if we can express it with kind of an
ever-increasing index. So the first thing that
might jump out at you is this oscillating sign that's
happening right over here. And whenever we see
an oscillating sign, that's a pretty good idea
that we could kind of view this as negative 1 to the
n-th, where n is our index. So for example, that
right over there is negative 1 to
the first power. That is-- this right
over here is negative 1 to the second power. That right over there is
negative 1 to the third power. So it looks like the
sign is being defined by raising negative
1 to the index. Now let's look at the
other parts of these terms right over here. So we have 5, then we
have 25, the we have 125, so these are the powers of 5. So this right over here
is 5 to the first power, this right over here is
5 to the second power, this right over here is
5 to the third power. So this part, we're
raising 5 to our index. Notice, 1, 1, 2, 2, 3, 3. And then finally,
let's look at this. We have 3, 6, and 9. So this literally-- if our index
here is 1, this is 3 times 1. If our index here is
2, this is 3 times 2. If our index here is
3, this is 3 times 3. So this is 3 times 1,
that is 3 times 2-- let me write it this
way-- 3 times 2, that right over
there is 3 times 3. So this sets us up
pretty well to write this in sigma notation. So let's write it over here
just so we can compare. So let me give myself some
real estate to work with. So we could write
this as the sum-- I'll do it in yellow-- as the
sum, so this is our sigma. We can start our index
n at 1, from n equals 1, and we're going to keep
going on and on forever. We just keep going
on and on forever. And so it's negative 1
to the n-th power times 5 to the n-th over-- notice
5 to the n-th-- over 3n is going to be equal to this. And you can verify
when n equals 1, it's negative 1
to the nth power-- I'm sorry-- negative 1
to the first power, which is negative 1 times 5 to
the first power, which is 5 over 3 times 1. And we can do that for
each successive term. And so we're all done.