Can the sum of an infinite geometric sequence be negative?
certainly
S = a/(1-r)
so, if a<0, S<0
extra credit: why cannot (1-r) be negative?
oobleck:
can you swap 1-r with r-1 if the common ratio is greater than one? In a question I was doing, the markscheme used r-1 and the sum was positive, but if you use 1-r the answer would be negative. Does it matter?
the problem is, that if 1-r < 0 then r > 1 and the series will not converge.
The formula still works for a finite number of terms, however.
Yes, the sum of an infinite geometric sequence can be negative.
To understand this, let's first explain what an infinite geometric sequence is. An infinite geometric sequence is a sequence of numbers where each term is found by multiplying the previous term by a constant ratio. For example, a geometric sequence with a first term of 1 and a common ratio of -1/2 would look like this: 1, -1/2, 1/4, -1/8, 1/16, ...
The sum of an infinite geometric sequence can be calculated using the formula: S = a / (1 - r), where "S" is the sum, "a" is the first term, and "r" is the common ratio. In the case of a geometric sequence with a negative common ratio, the sum can indeed be negative.
For example, let's consider the geometric sequence with a first term of 1 and a common ratio of -1/2. Using the formula, we can calculate the sum as follows:
S = 1 / (1 - (-1/2))
= 1 / (1 + 1/2)
= 1 / (3/2)
= 2/3
So, the sum of this infinite geometric sequence is 2/3, which is positive. However, if we look at the terms of the sequence, we can see that they alternate between positive and negative values, resulting in a negative partial sum before approaching 2/3 as the limit of the sequence.