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
(-1)^k
---------
3^(k+1)
-1/9 , 1/27 , -1/3^4 ...
yes, g = -1/9
no r = -1/3 (disagree with your sign)
Sum = -(1/9 ) / (1 +1/3) =-1/12
To determine if the series is convergent, you can use the formula for the sum of a geometric series. The formula is:
S = a / (1 - r)
where:
S represents the sum of the series,
a is the first term of the series, and
r is the common ratio of successive terms.
In this case, the first term (a) is -1/9 and the common ratio (r) is 1/3. Substituting these values into the formula, we can calculate the sum:
S = (-1/9) / (1 - 1/3)
Note that the absolute value of the common ratio (r) must be less than 1 for the series to converge. In this case, 1/3 satisfies that condition.
Simplifying the expression, we get:
S = (-1/9) / (2/3)
S = (-1/9) * (3/2)
S = -1/6
Therefore, the series converges to -1/6, not 1/6 as you stated.