Another problem:
determine whether the series is convergent if so find sum
it is the sum from k=1 to infinity of ((-1)^k)/(3^(k+1))
i found this series to be geometric where
a=-1/9 and
r=1/3
my answer was converges to 1/6
To determine whether the series is convergent, we can use the formula for the sum of an infinite geometric series. The formula is:
S = a / (1 - r)
where "S" is the sum of the series, "a" is the first term, and "r" is the common ratio.
In this case, the first term "a" is -1/9 and the common ratio "r" is 1/3.
Now, let's plug these values into the formula:
S = -(1/9) / (1 - 1/3)
Simplifying:
S = -(1/9) / (2/3)
To divide by a fraction, we can multiply by its reciprocal:
S = -(1/9) * (3/2)
Simplifying:
S = -1/6
Therefore, the sum of the series from k=1 to infinity of ((-1)^k)/(3^(k+1)) is -1/6.
So, you were correct in determining that the series converges to -1/6, not 1/6.