does the sum 3/(2^(n)-1) converge or diverge?
n=1--->infinity
To determine whether the sum 3/(2^(n)-1) converges or diverges, we can analyze its behavior as n approaches infinity.
First, let's rewrite the series by expanding the denominator using the formula for geometric series:
3/(2^(n)-1) = 3/(2^(n)) * 1/(1 - 1/2^(n))
Now, let's simplify this expression further:
3/(2^(n)) * 1/(1 - 1/2^(n)) = 3 * (1/2^(n)) / (1 - 1/2^(n))
Now, as n approaches infinity, 1/2^(n) becomes very close to 0. When we divide 1 by a number that is very close to 0, the result tends to infinity. Therefore, the denominator in the expression above approaches zero as n approaches infinity.
To determine if the sum converges or diverges, we need to examine the behavior of the numerator. In this case, the numerator is just the constant 3.
Since the denominator approaches 0 and the numerator remains constant, the overall expression tends to infinity as n approaches infinity. Therefore, the sum 3/(2^(n)-1) diverges as n goes from 1 to infinity.