Find the sum of the infinite series 1/3+1/9+1/27+... if it exists
this is a GS where
a = 1/3
r =1/3
sum(infinite # of terms) = a/(1-r)
= (1/3)/(1 - 1/3)
= (1/3) / (2/3)
= (1/3)(3/2)
= 1/2
To find the sum of an infinite series, we need to determine if the series converges or diverges. Let's analyze the given series: 1/3 + 1/9 + 1/27 + ...
This series is a geometric series with a common ratio of 1/3. To determine if it converges, we can use the formula for the sum of an infinite geometric series, given by:
Sum = a / (1 - r),
where "a" is the first term of the series and "r" is the common ratio.
In this case, the first term "a" is 1/3 and the common ratio "r" is also 1/3. Plugging these values into the formula, we have:
Sum = (1/3) / (1 - 1/3).
Simplifying this expression, we get:
Sum = (1/3) / (2/3) = (1/3)*(3/2) = 1/2.
Therefore, the sum of the infinite series 1/3 + 1/9 + 1/27 + ... exists and is equal to 1/2.