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Series

Module by: Kenny M. Felder. E-mail the author

Summary: An updated version of the Series module.

A series is a list of numbers—like a sequence—but instead of listing them, you add them all up. For instance, 4+9+3+2+17. (This particular series adds up to 35.)

One way to compactly represent a series is with “summation notation,” which looks like this:

n = 3 7 n 2 n = 3 7 n 2 size 12{ Sum cSub { size 8{n=3} } cSup { size 8{7} } {n rSup { size 8{2} } } } {}
(1)

The big funny-looking thing in the middle is the Greek letter uppercase Sigma, and it indicates a series. To “unpack” this notation, start counting at the bottom ( n = 3 n=3), and stop when you reach the stop ( n = 7 n=7). For each term, plug that value of n n into the given formula ( n 2 n 2 ). So this particular formula, which we can read as “the sum as n n goes from 3 to 7 of n 2 n 2 ”, simply means:

3 2 + 4 2 + 5 2 + 6 2 + 7 2 3 2 + 4 2 + 5 2 + 6 2 + 7 2

Arithmetic Series

If you add up all the terms of an arithmetic sequence, you have an arithmetic series. For instance, 10 + 13 + 16 + 19 + 22 + 25 = 105 10+13+16+19+22+25=105.

There is a “trick” that can be used to add up the terms of any arithmetic series. While this trick may not save much time with a 6-item series like the one above, it can be very useful if adding up longer series. The trick is to work from the outside in.

Consider the example given above: 10 + 13 + 16 + 19 + 22 + 25 10+13+16+19+22+25. Looking at the first and last terms, 10 + 25 = 35 10+25=35. Going in, to the second and next-to-last terms, 13 + 22 = 35 13+22=35. Finally, the two inside numbers 16 + 19 = 35 16+19=35. So we can see that the sum of the whole thing is 3 * 35 3*35.

Pause here and check the following things.

  • You understand the calculation that was done for this particular example.
  • You understand that this “trick” will work for any arithmetic series.
  • You understand that this trick will not work, in general, for series that are not arithmetic.

If we apply this trick to the generic arithmetic series, we get a formula that can be used to sum up any arithmetic series.

Every arithmetic series can be written as follows:

t 1 + ( t 1 + d ) + ( t 1 + 2 d ) ( t n - d ) + t n t 1 +( t 1 +d)+( t 1 +2d)( t n -d)+ t n

If you add the first and last terms, you get t 1 + t n t 1 + t n . Ditto for the second and next-to-last terms, and so on. How many such pairs will there be in the whole series? Well, there are n n terms, so there are n2n2 size 12{ { {n} over {2} } } {} pairs. So the sum for the whole series is n2( t 1 + t n ) n2 size 12{ { {n} over {2} } } {}( t 1 + t n ).

Geometric Series

If you add up all the terms of a geometric sequence, you have a geometric series. The “arithmetic series trick” will not work on such a series; however, there is a different trick we can use. As an example, let’s find the sum 2 + 6 + 18 + 54 + 162 2+6+18+54+162.

We begin by calling the sum of this series S S:

S = 2 + 6 + 18 + 54 + 162 S=2+6+18+54+162

Now, if you multiply both sides of this equation by 3, you get the first equation I have written below. (The second equation below is just copied from above.)

3 S = 6 + 18 + 54 + 162 + 486 3S=6+18+54+162+486 (*confirm this for yourself!)

S = 2 + 6 + 18 + 54 + 162 S=2+6+18+54+162

Here comes the key moment in the trick: subtract the two equations. This leaves you with:

2 S = 486 - 2 2S=486-2, so S = 242 S=242

Once again, pause to convince yourself that this will work on all geometric series, but only on geometric series.

Finally—once again—we can apply this trick to the generic geometric series to find a formula. So we begin with t 1 + t 1 r + t 1 r 2 + t 1 r 3 t 1 r n - 1 t 1 + t 1 r+ t 1 r 2 + t 1 r 3 t 1 r n - 1 and write…

r S = t 1 r + t 1 r 2 + t 1 r 3 + t 1 r 4 ... t 1 r n - 1 + t r r n rS= t 1 r+ t 1 r 2 + t 1 r 3 + t 1 r 4 ... t 1 r n - 1 + t r r n (*confirm this!)

S = t 1 + t 1 r + t 1 r 2 + t 1 r 3 + ... t 1 r n - 1 S= t 1 + t 1 r+ t 1 r 2 + t 1 r 3 + ... t 1 r n - 1

Again, subtracting and solving, we get…

r S S = t 1 r n t 1 rSS= t 1 r n t 1

S ( r - 1 ) = t 1 ( r n –1 ) S(r-1)= t 1 ( r n –1)

S = t 1 rn1r1S= t 1 rn1r1 size 12{ { {r rSup { size 8{n} } - 1} over {r - 1} } } {}

So there we have it: a general formula for the sum of any finite geometric series, with the first term t 1 t 1 , the common ratio r r, and a total of n n terms.

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