please help me...for the power series summation of n^4x^n/2(n-1)factorial....find the radius
and interval of convergence...
so far i am guessing that radius is equal to x and interval of convergence is (-infinty to positive infinity) is that correct?
You need to do a ratio test first. Have you done that?
no i didnt...how would ratio test help me?
K I DID USE THE RATIO TEST and i got the limit equal to 0...so does that mean that the radius of convergence is equal to infinity? so is the radius x?
For what values of x does it converge in the ratio test? Doesn't that determine the radius of convergence?
Yes, you're right! The ratio test is a tool that helps us determine the radius and interval of convergence for a power series. When using the ratio test, we compare the absolute value of consecutive terms in the series and take the limit as n approaches infinity.
To apply the ratio test, let's consider the series you provided: ∑ (n^4 * x^n) / (2(n-1) factorial).
First, let's write the expression for the ratio test:
L = lim(n->∞) |(a_(n+1)) / (a_n)|,
where a_n represents the n-th term in the series.
Next, let's substitute the expression for a_n and a_(n+1):
L = lim(n->∞) |[(n+1)^4 * x^(n+1)] / (2n factorial) * (2(n-1) factorial) / (n^4 * x^n)|.
Now, we can simplify the expression:
L = lim(n->∞) |(n+1)^4 * x^(n+1) * n^4 * x^n / [2n! * 2(n-1)!]|.
L = lim(n->∞) |(n+1)^4 * x * x^n * n^4 / (2n! * 2(n-1)!)|.
L = |x| * lim(n->∞) |(n+1)^4 * n^4 / (2n! * 2(n-1)!)|.
Now, let's simplify the expression inside the limit:
L = |x| * lim(n->∞) [(n^4 + 4n^3 + 6n^2 + 4n + 1) * n^4] / (2n! * 2(n-1)!).
L = |x| * lim(n->∞) [(n^8 + 4n^7 + 6n^6 + 4n^5 + n^4) / (2n! * 2(n-1)!)]
L = |x| * lim(n->∞) [(n^8 + 4n^7 + 6n^6 + 4n^5 + n^4) / (2n * (2n-1) * 2(n-1)!)],
where we used the fact that (2n)! = 2n * (2n-1) * (2n-2) * ... * 2 * 1.
Now, let's simplify further:
L = |x| * lim(n->∞) (n^8 + 4n^7 + 6n^6 + 4n^5 + n^4) / (2n * (2n-1) * 2(n-1)!).
As n goes to infinity, all the terms except n^8 will become negligible. So, we can ignore them:
L = |x| * lim(n->∞) n^8 / (2n * (2n-1) * 2(n-1)!).
Now, we can further simplify the expression inside the limit:
L = |x| * lim(n->∞) n^7 / ((2n-1) * (2n-2)!).
Notice that the limit above is equal to infinity. Therefore, regardless of the value of x, L is always infinity.
According to the ratio test, if L is equal to infinity, the series diverges for all x ≠ 0. Thus, the radius of convergence is 0.
In other words, the series converges only at x = 0, and it is not convergent for any other value of x.
Hence, the interval of convergence for this series is the singleton set {0}.