If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### 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!