Consider the series 1/4+1/6+1/9+2/27+4/81+....
Does the series converge or diverge? Is the series arithmetic, geometric, neither, or geometric with an absolute value of the common ration being greater/ less than 1?
3 answers
since r = 2/3 it will converge
Does the series converge or diverge? If it converges, what is the sum? Show your work.
∑∞n=1−4(−1/3)n−1
∑∞n=1−4(−1/3)n−1
This is a geometric series with first term a=-4 and common ratio r=-1/3. In order for it to converge, we need to have |r| < 1, which is the case here.
The sum of an infinite geometric series with first term a and common ratio r, where |r| < 1, is given by:
S = a / (1-r)
Plugging in the values a=-4 and r=-1/3, we get:
S = (-4) / [1 - (-1/3)]
S = (-4) / (4/3)
S = -3
Therefore, the series converges and its sum is -3.
The sum of an infinite geometric series with first term a and common ratio r, where |r| < 1, is given by:
S = a / (1-r)
Plugging in the values a=-4 and r=-1/3, we get:
S = (-4) / [1 - (-1/3)]
S = (-4) / (4/3)
S = -3
Therefore, the series converges and its sum is -3.
2 answers