The sum from 0 to infinity of (-1)^n(3/3^n) is convergent or divergent? If convergent, what is the sum?
I got that it's convergent and the sum is 9/2, but that's wrong.
It is not clear without sufficient parentheses what the expression really is.
I assume it to be:
Sum((-1)^n * (3/3^n)) for n=0 -> ∞
This is an alternating geometric series.
(9/2) is the correct sum for the geometric series (non-alternating).
Write out the first few terms of the series:
3 - 3/3 + 3/9 - 3/27 + 3/81 - ...
which can be regrouped into two geometric series:
3(1+1/9+1/81+...) - (1+1/9+1/81+...)
=3(9/8) - (9/8)
=2(9/8)
=9/4
Note that if the minus sign becomes a plus sign, we get the geometric sum of 9/2.
Oh, right . I forgot all about the -1. Thanks.
You're welcome!
To determine whether the sum from 0 to infinity of (-1)^n(3/3^n) is convergent or divergent, we can use the concept of geometric series.
A geometric series has the form: a + ar + ar^2 + ar^3 + ... , where 'a' is the first term and 'r' is the common ratio between consecutive terms.
In this case, our first term is (-1)^0(3/3^0) = 1(3) = 3. The common ratio is the ratio between consecutive terms, which in this case is (3/3) = 1.
A geometric series is convergent if the absolute value of the common ratio, |r|, is less than 1. In our case, |r| = 1. Since |r| equals 1, the geometric series will be divergent.
Therefore, the sum from 0 to infinity of (-1)^n(3/3^n) is divergent.