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## Convergent and divergent infinite series

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# Partial sums: formula for nth term from partial sum

AP.CALC:

LIM‑7 (EU)

, LIM‑7.A (LO)

, LIM‑7.A.1 (EK)

, LIM‑7.A.2 (EK)

## Video transcript

- [Narrator] Nth partial
sum of the series, we're going from one
to infinity, summing it a sub n is given by. And they tell us of the formula for the sum of the first n terms. And they say write a rule for what the actual Nth term is going to be. Now to help us with this, let me just create a little visualization here. So if I have a sub one
plus, plus a sub two, plus a sub three, and I keep adding all the way to a sub n minus one plus a sub n. This whole thing, this whole thing that I just wrote out. That is s sub n. This whole thing is s, let me this whole thing is sub n, which is equal to n plus one, over n plus 10. Now if I wanna figure out a sub n, which is the goal of this exercise, well I could subtract out the sum of the first n minus one terms. So I could subtract out this. So that is s, that is s sub n minus one, and what would that be equal to. Well wherever we see
an n, we'd replace with an n minus one. So it'd be n minus one plus one over n minus one plus 10, which is equal to n over n plus nine. So if you subtract the red
stuff from the blue stuff, all that you're gonna be
left with is the thing that we're gonna solve for. You're gonna be left with a sub n. So we could write down a
sub n is equal to s sub n, is equal to s sub n
minus s sub n minus one S sub n minus one. Or we could write, that
is equal to this stuff. So this is the n plus one over n plus 10 minus, minus n over n plus nine. And this by itself, this
is a rule for a sub n. But we could combine these terms, add these two fractions together. And this is actually going to be the case for n greater than one. For n equals one, s
sub one is going to be, well you can just, a sub
one is going to be equal to s sub one. But then for any other
n, we could use this right over here. And if we want to
simplify this, well we can add these two fractions. We can add these two fractions by having a common denominator. So let's see, if we multiply the numerator and denominator
here times n plus nine, we are going to get, this is equal to n plus
one times n plus nine, over n plus 10 times n plus nine. And from that, we are going to subtract. Let's multiply the
numerator and denominator here by n plus 10. So we have n times n plus 10, over n plus nine times n plus 10. N plus nine times n plus 10. And what does that give us? So, if we simplify up
here, we're gonna have, this is n squared plus 10
n plus nine, that's that. And then this right over
here this is n squared plus 10 n, within that red color. So this is n squared plus 10 n and remember we're gonna subtract this. And so, and we are close
to deserving a drum roll, a sub n is going to be
equal to our denominator right over here is n plus
nine times n plus 10. And we're gonna subtract the
red stuff from the blue stuff. So you subtract an n
squared from an n squared and those cancel out. Subtract a 10 n from a
10 n, those cancel out. And you're just left with that blue nine. So there you have it, we've expressed, we've written a rule for a
sub n, for n greater than one.