find the interval of convergence of the sum of ((x^1/2)/2 - 1)^k from k = 0 to infinity and within this interval find the sum of the series as a function of x.
I know the interval is 0 < x < 16 but how do I get the sum?
To find the interval of convergence of the given series and determine the sum as a function of x, you can use the ratio test. Let's break down the steps:
Step 1: Use the ratio test.
The ratio test states that for a series ∑aₙ to converge, if the limit of |aₙ₊₁ / aₙ| as n approaches infinity is less than 1, the series converges. Conversely, if the limit is greater than 1 or does not exist, the series diverges.
In our case, aₙ = ((x^(1/2))/2 - 1)^k. Let's calculate aₙ₊₁ / aₙ:
aₙ₊₁ / aₙ = (((x^(1/2))/2 - 1)^(k+1)) / (((x^(1/2))/2 - 1)^k)
= ((x^(1/2))/2 - 1)^(k+1 - k) [since (a^b)^c equals a^(b*c)]
= ((x^(1/2))/2 - 1).
Step 2: Find the limit of ((x^(1/2))/2 - 1) as n approaches infinity:
To determine the interval of convergence, we need to find the values of x for which |((x^(1/2))/2 - 1)| < 1.
|((x^(1/2))/2 - 1)| < 1 leads to:
-1 < ((x^(1/2))/2 - 1) < 1
0 < (x^(1/2))/2 < 2
0 < x^(1/2) < 4
0 < x < 16.
Therefore, the interval of convergence is 0 < x < 16.
Step 3: Determine the sum as a function of x.
Now that we know the interval of convergence, we need to find the sum of the series within this interval. The sum can be given as a function of x by applying the formula for the sum of a geometric series:
S = a / (1 - r),
where S is the sum, a is the first term, and r is the common ratio.
In our case, the given series starts from k = 0. Therefore, the first term (a) is ((x^(1/2))/2 - 1)^0, which simplifies to 1. The common ratio (r) is ((x^(1/2))/2 - 1).
Thus, the sum of the series as a function of x is:
S = 1 / (1 - ((x^(1/2))/2 - 1)).
Simplifying further:
S = 1 / (3/2 - (x^(1/2))/2).
So, the sum of the series as a function of x is 1 / (3/2 - (x^(1/2))/2).