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### Course: Integral Calculus>Unit 5

Lesson 16: Representing functions as power series

# Converting explicit series terms to summation notation

Some sums are really long, but we can use sigma notation to write them in a shorthand way. See such an example here. Created by Sal Khan.

## Want to join the conversation?

• Wouldn't (-5)^n/3n work just as well as (-1)^n(5^n/3n)? Is there any particular reason to favor one form over the other?
• Your expression works just as well and is more compact. I'm guessing Sal chose this way to highlight the use of -1 to a power as a way to create an alternation between positive and negative values as a distinct technique that can be used regardless of whether your expression includes another value that's being raised to the same power.
• Would (-5)^(n+1)/3(n+1) work just as well but instead of n=1 I would start at n=0 for the notation.......?
• Yes, that's correct. The ability to see that sigma notation can be manipulated this way shows you understand the notation. Generally, though, we look to have the index begin at a number that gives us a simpler expression to the right of the sigma, so we'd prefer to see n instead of n + 1 as an exponent and multiplicand. Sort of like simplifying a fraction before giving the final answer.
• Could someone explain to me why 5 is raised to the nth power but 3 is only multiplied by it?
• The terms in the numerator are exponential powers of 5: 5^1 = 5, 5^2 = 25, 5^3 = 125 ; the terms in the denominator are multiples of 3: 3*1 = 3, 3*2 = 6, 3*3 = 9.
• What if the series isn't a fraction?
• The notation doesn't depend on the type of numbers. You just analyse the series and then note the sigma notation.
• Is the series featured in the video geometric?
• No, it isn't; I've never seen a particular name give to this kind of series. The idea here is to find the nth term so that you can manipulate this equation however the problem requires.
• Why does the bottom of sigma, i=1 equal 1 rather than the first number in the sequence -5/3?
• The `i` is the index value of the series (the element's position in the sequence of which this is a series). It is not the actual value of the element, just its position. As such, it is numbered as an integer beginning at one or sometimes at 0.
• i wrote mines as ((-5)^k)/3k when k starts with 1 and if k starts with 0 [(-5(-5)^k)/3+3k]
• If you write out the first 5 terms in each series they should be the same.
It is usually best to separate the term (-1)^k or (-1)^(k+1) to determine the sign of the terms.
• Please let me know if I am on the right track...
5/2 -5/4 +5/6 -5/8 +....

Is the sigma notation equation = (-1)^n (5^1/n+2)
(1 vote)
• Take another look at the n + 2 term in your denominator. Does it match up with terms in the sequence for a given n? Also, make sure you know where n is starting from. Are you saying that it starts as 1, or 0? Either way is fine, but you have to make sure you're consistent throughout your expression and sometimes one way is easier for a particular problem. Hope this helps!
• I was just wondering on how it would be written for example if the difference between each term is different, for instant the series 5 + 10 + 17 + 16 (Difference is +2 each time, how would you write that?
(1 vote)
• This would be a quadratic, because the difference of the differences is constant.
• Could we also use: (5^(-n))/(3n) or would that be incorrect?
(1 vote)
• That would be incorrect, since 5^(-n) is equal to (1/5)^n, not (-5)^n. You could use (-5)^n, though, because (-1)^n * 5^n is equiavlent to (-1 * 5)^n, or (-5)^n.
(1 vote)

## 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.