Question

What is a sample problem about finding the sum of a geometric sequence?

Answer 1

A geometric sequence is a sequence that involves multiplication to achieve a sequence. It is defined by:

`a_n = a_0r^(n-1)` `AA` `n>=1` and `a_0 = a'`,
where `a'` is some arbitrary first number or function and `r` is some constant multiplier.

If you applied this to one like this:
`1, 2, 4, 8, ...`
you'd get:

`a_n = 1*2^(n-1)` `AA` `n>=1`, `a_0 = 1`, and `r = 2`.

A fuller example...
If you looked at this:
`1, 1/2, 1/4, 1/8, 1/16, ...`
you can see that it's multiplying by `1/2`.

`a_n = 1*(1/2)^(n-1)` `AA` `n>=1`, `a_0 = 1`, and `r = 2`.

`AA` means "for all [in the set]".

Now if you wanted to write out a sum, you could begin with the power series:
`sum_(n=0)^N x^n = 1 + x + x^2 + x^3 + ...`
which happens to equal:
`1/(1-x)`

Then you can substitute in `1/2` since `1/2` was `r`:

`=> 1 + (1/2) + (1/2)^2 + (1/2)^3 + ...`
`=> 1 + 1/2 + 1/4 + 1/8 + 1/16 + ...`
`=>1/(1-1/2) = 1/(1/2) = 2`

So it sums, and converges to 2. This, by the way, proves Zeno's Paradox false. You can take an infinite number of steps with halved distances and travel twice your initial distance.

If you had this instead:
`1 - 1/2 + 1/4 - 1/8 + 1/16 - ...`

You just have:
`sum_(n=0)^N (-1)^nx^n = sum_(n=0)^N (-x)^n = 1 - x + x^2 - x^3 + ...`

and so the sum is:
`1/(1-(-x)) = 1/(1+x)`

And here, the sum is this if x = `1/2`:
`1/(1-(-1/2)) = 1/(3/2) = 2/3`