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### Course: 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.

## Want to join the conversation?

- I came up with the formula (-1)^(n-1) * 8(n-1)/(2n+1). Is this correct as well?(2 votes)
- Not quite. Your formula fails once we reach a_4, where Sal has a term of -32/9, but your function would give (-1)^(4-1) * 8(4-1)/(2*4 + 1) = -24/9. Yours would work if the numerator were simply increasing in magnitude by 8 each time, but here's it actually doubling. Hope that helps!(8 votes)

- Hey, so I just figured out that the common ratio between these successive terms is -10/7, therefore shouldn't the fourth term be -160/49 instead of -32/9?(1 vote)
- Well... In a word, no. The ratio between the first two terms may be -10/7, but the fourth term is part of the series as Sal has defined it, too, so we can't just discount it or say that it must be wrong when we're trying to figure out the pattern. This is not a geometric series, but if you just look at the first two terms, you might think it is. In fact, if you just look at the first two terms of
**any**series, you could convince yourself that it's geometric because there will always be some constant ratio between two given numbers.

If we knew that this series was geometric, your statement would be correct; however, we don't know that, and in fact it's not a geometric series, so we can't assume that there will be a common ratio. Hope that makes sense.(7 votes)

- what does a sub n mean?(2 votes)
- a sub n, which people often format as a_n is the last number in a series, or any particular number represented by its position n, in the a series (a_1, a_2, a_3, ..., a_n). If we have 10 numbers in some series, the tenth number, a_10, will be the a_n. In the example in this video, the first term is
`a_2`

. And`a_n`

would be an infinite unkwown number. Bytheway, there are sums that the infinite sum of n numbers is known, like`sum of 2^-n, n=0 to infinity`

. http://www.wolframalpha.com/input/?i=sum+of+2%5E-n,+n%3D0+to+infinity(1 vote)

- How can I identify patterns in series easier? I know it's broad, but that's my main problem when completing these questions. Once I identify the pattern, I can come up with the general term pretty easy.(2 votes)
- Look for the usual suspects: repetition, distance between terms, ratio between terms, alternation patterns. The best method is just to practice it. Look for other resources online, calculus texts, etc. Remember way back when factoring a quadratic seemed difficult, but now, with lots of experience under your belt, no problems, right? Same deal here.(1 vote)

- i have a different formula! is this right [(-2)^n * 2] / -(2n +1)(2 votes)
- Can i write it as 4*(-2)^n / 3+2n instead?(1 vote)
- There are an infinite number of ways to express most sequences/series explicitly. Your expression is close; however, at n = 1 your expression = -8/5. If you look at Sal's expression you will notice his series starts off with n = 2. Thus, if you were to decrease the index of your expression by one you would get a correct answer as well. For example,

4*(-2)^(n-1) / 3+2(n-1)

4*(-2)^(n-1) / 3+2n-2

4*(-2)^(n-1) / 2n+ 1

Notice, if you manipulated the final expression above enough, you would get same answer as Sal.

(Focusing only on the numerator)

4 = 2*2

(-2)^(n-1) = (-2)^n / -2

2*2*(-2)^n / -2

2 and -2 cancel out to equal -1

-1*2*(-2)^n

-1 * 2 = -2

-2*(-2)^n = (-2)^(n+1)

Thus, your expression = (-2)^(n+1) / 2n+1(2 votes)

- Could another Sigma notation for this same problem be: (-5)^n/3n?

My reasoning is, (-5)^1 = -5, ^2 = 25, ^3 = -125, etc, and it would be much simpler. Would that work as a solution for this as well? Thanks! Sorry, I'm watching the video "Writing a series in Sigma notation" but the comments are from a different video. Not sure what's going on with the site, but I went back and checked other videos and the same thing is happening there. I'll try to put the comment on the right video.(1 vote)- Yes, that would work; I guess you couldn't see the video comments, but that idea was at the top. And actually, you could even go so far as to express it as (-5/3)^n, which I think may be even more useful because it really shows the geometric nature of this series. As another user said, I suspect Sal used the (-1)^n notation because he wanted to use the concept of an oscillator, which is a good thing to be able to recognize and apply.(2 votes)

- I know that this sequence may not be completely simplified, but, for sigma terms, is it okay if i do 4*(-2n^n-1)/3+2(n-1)?(1 vote)
- Hey, So could this series be explicitly defined with another formula? Because I came up with another formula, different from yours, and mine works too.(1 vote)
- In the previous video (Writing a series in sigma notation) Sal takes great pains to use (-1)^n for alternating the sign of the terms, now 5 minutes later in video time he's completely ignoring that concept. Why did he not say (-1)^(n-1) *2^(n+1)/(2n+1) to be consistent?(1 vote)
- It can be done either way. Separating out the (-1)^n term is akin to writing 7 - 6 as

7 + -6. Once you get comfortable with the notation, either is easily recognizable. Wrapping the negative into the exponential base gives you a simpler expression, though.(1 vote)

## 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.