Find the radius of convergence, R, of the series.
n=1 ((-1)^n x^n) /n^6
To find the radius of convergence, we can use the ratio test:
|r| = lim|n→∞| |-x(n+1) / ((n+1)^6)| / |-x^n / (n^6)|
Simplifying, we get:
|r| = lim|n→∞| |x(n+1)/x^n| * |n^6 / (n+1)^6|
Since |-x(n+1) / x^n| = |x| for all n, we can ignore that factor. Now we just need to find the limit of the second factor:
lim|n→∞| n^6 / (n+1)^6
Using L'Hopital's rule, we can evaluate this limit as:
lim|n→∞| 6n^5 / 6(n+1)^5 = lim|n→∞| (n / n+1)^5 = 1
So the radius of convergence is:
|r| = |x| * 1 = |x|
Therefore, the radius of convergence is R = |x|.
what would be the intervals of convergence?
To find the interval(s) of convergence, we need to check the endpoints of the interval [-R, R] to see if the series converges or diverges at those points.
At x = -R, the series becomes:
∑((-1)^n*(-R)^n)/n^6
Using the alternating series test, we can see that this series converges.
At x = R, the series becomes:
∑((-1)^n*R^n)/n^6
This series is not an alternating series and we can't use the alternating series test. However, we can use Leibniz's test for convergence of alternating series with decreasing terms. We need to show that the terms of the series decrease in absolute value to 0 as n increases.
We can do this by comparing the series to the series with the absolute values of the terms:
∑(R^n)/n^6
Using the ratio test, we can see that this series converges. Therefore, by the comparison test, we can conclude that the series at x = R also converges.
Since the series converges at the endpoints of the interval [-R, R], and we have shown that the series converges absolutely for all values of x in this interval, we can conclude that the interval of convergence is:
[-R, R] = [-|x|, |x|].
Therefore, in this case, the interval of convergence is symmetric about the origin.
To find the radius of convergence, R, of the given series, we can use the Ratio Test. The Ratio Test states that for a series with terms a_n, the series converges if the limit of |a_(n+1) / a_n| as n approaches infinity is less than 1, and it diverges if the limit is greater than 1.
In this case, the series is given by a_n = ((-1)^n x^n) / n^6.
Let's apply the Ratio Test to determine the convergence of the series.
First, let's compute a_(n+1) / a_n:
a_(n+1) / a_n = [((-1)^(n+1) x^(n+1)) / (n+1)^6] / [((-1)^n x^n) / n^6]
= [((-1)^(n+1) x^(n+1)) / (n+1)^6] * [n^6 / ((-1)^n x^n)]
We can simplify this expression by canceling out common factors between the numerator and denominator:
= (-1) * (x^(n+1)) * (n^6) / (n^6 * (n+1)^6 * ((-1)^n x^n))
= -x(n+1) / (n+1)^6 = -x/(n+1)^5
Now, let's take the absolute value of the above expression:
|a_(n+1) / a_n| = |(-x/(n+1)^5)| = |x/(n+1)^5|
Since the absolute value of x does not affect convergence, we can ignore it for now.
Now, let's take the limit of |a_(n+1) / a_n| as n approaches infinity:
lim(n->∞) |x/(n+1)^5|
As n approaches infinity, the denominator, (n+1)^5, becomes very large compared to the numerator, x. Therefore, the entire fraction approaches 0.
lim(n->∞) |x/(n+1)^5| = 0
Since the limit is less than 1, the series converges for any value of x. Hence, the radius of convergence, R, is infinite for the given series.