Find the R and interval
summation from n=1 to infinity n/bn (x − a)^n, b > 0
what's with the n/bn ?
n over b^n times (x − a)^n
nvm i got it
For some good examples and explanation, see
https://tutorial.math.lamar.edu/classes/calcii/PowerSeries.aspx
To find the radius of convergence (R) and the interval of convergence, we can use the ratio test. The ratio test states that for a power series ∑(n=1 to ∞) cnx^n, if the limit of the absolute value of the ratio of consecutive terms as n approaches infinity, lim(n→∞) |(cn+1x^(n+1))/(cnx^n)|, exists and is equal to L, then the power series converges absolutely if L < 1 and diverges if L > 1.
Let's apply the ratio test to the given series ∑(n=1 to ∞) n/bn (x − a)^n.
First, let's simplify the series by factoring out the common terms inside the sum:
∑(n=1 to ∞) (n/bn) (x − a)^n = ∑(n=1 to ∞) [n(x-a)/b]^n
Applying the ratio test:
lim(n→∞) |(n+1(x-a)/b) / (n(x-a)/b)| = lim(n→∞) |(n+1)(x-a)/n(x-a)|
To simplify the limit, we can cancel out the common (x-a) terms:
lim(n→∞) |(n+1)/n|
Now, let's evaluate the limit:
lim(n→∞) |(n+1)/n| = 1
Since the limit is equal to 1, we can conclude that the ratio test is inconclusive and cannot determine the convergence or divergence of the series. In this case, we need to use another convergence test.
To find the radius of convergence (R) and the interval of convergence, we can use the root test. The root test states that for a power series ∑(n=1 to ∞) cnx^n, if the limit of the nth root of the absolute value of the terms as n approaches infinity, lim(n→∞) (|cnx^n|)^(1/n), exists and is equal to L, then the power series converges absolutely if L < 1 and diverges if L > 1.
Let's apply the root test to the given series ∑(n=1 to ∞) n/bn (x − a)^n.
Using the nth root:
lim(n→∞) ((n/bn) (x − a)^n)^(1/n) = lim(n→∞) ((n/bn)^(1/n)) ((x − a)^n)^(1/n)
For the first part of the limit, ((n/bn)^(1/n)), we can use the fact that lim(n→∞) (1/n) = 0:
lim(n→∞) ((n/bn)^(1/n)) = (1/1)^0 = 1
For the second part of the limit, ((x − a)^n)^(1/n), we can simplify it as (x - a):
lim(n→∞) ((x − a)^n)^(1/n) = lim(n→∞) (x - a) = x - a
Multiplying the results from the two parts of the limit:
lim(n→∞) ((n/bn) (x − a)^n)^(1/n) = 1 * (x - a) = x - a
Now, based on the root test, for the series to converge absolutely, the result of the limit must be less than 1. Therefore, we have:
x - a < 1
To find the interval of convergence, we also need to consider the endpoints. Substituting x - a = 1, we have:
1 < 1
which is not true.
Therefore, we can conclude that the series does not converge absolutely at x = a + 1.
For the other endpoint, we substitute x - a = -1:
-1 - a < 1
-2 - a < 0
a > -2
Thus, the interval of convergence is (a, a + 1], which means the series converges absolutely for all values of x within this interval, excluding the endpoint x = a + 1.
Note: The value of b does not affect the radius of convergence or the interval of convergence in this case, as it only appears in the series as a constant coefficient.