Calculus: Find the interval of convergence of the power series: (x-1)^n/3^n from n=0 to infinity?

Answer 1

Take absolute value and apply the ratio test

a_n+1/a-n= Ix-1I *1/3 whose limit is
Ix-1I/3
If Ix-1I/3<1 the series is absolutely convergent
-3 a_n=1 divergent and at x=-2 an =(-1)^n also divergent

Answer 2

3

Answer 3

When you want to find the interval of convergence, you take the limit as n goes to infinity of abs(An+1/An), as if using the ratio test. You then set this value less than 1 and solve for x. This x value is your radius of convergence, then you test the endpoints to find the interval of convergence.

In this case, then, we compute (I've gone ahead and multiplied by the reciprocal):
lim|((x-1)^(n+1)*(3^n))/(3^(n+1)*(x-1)^n)|
lim|(x-1)/3|
(1/3)|x-1| < 1
|x-1| < 3
-3 < x-1 < 3
-2 < x < 4

Now you plug these two endpoints into your original power series to see if they make it converge or diverge. In this case you get -1 or 1 for your power series, neither of which converges, so the answer is (-2, 4).

Answer 4

to discover the era of convergence |x-a| < R, use the ratio or root attempt (commonly, ratio attempt will yield the ensuing era). utilising the ratio attempt: a million. lim_n-inf._[ | [(x-8)^(n+a million) / 8^(n+a million)] * [8^n / (x-8)^(n)] | ]. 2. Simplify the expression [(x-8)^(n+a million) / 8^(n+a million)] * [8^n / (x-8)^(n)] interior the shrink: [(x-8)^n * (x-8)^a million * 8^n] / [(x-8)^n * 8^n * 8^a million]; (x-8)^n term and eight^n term cancel leaving: (x-8)/8. 3. The shrink simplifies to lim_n-inf._[ | (x-8)/8| ]. by using fact the expression does not remember on 'n', you are able to handle it as a relentless. utilising the shrink-consistent rule, lim_n-inf._[ (x-8)/8| ] = |(x-8)/8|. 4. in accordance to the ratio attempt for the sum of a team A_n from n = a million to infinity, if lim_n-inf._( |A_n+a million / A_n| ) < a million, then the sequence converges truthfully (If the sequence converges truthfully, there's a theorem that publicizes that the sequence additionally converges). If lim_n-inf._( |A_n+a million / A_n |) > a million or equals infinity or DNE, then the sequence diverges. subsequently, the sequence with the series (x-8)^n / 8^n converges if |(x-8)/8| < a million and diverges otherwise. 5. discover the radius of convergence by potential of manipulating the inequality you ended with after taking the shrink which consequently is (x-8)/8 < a million. a. |(x-8)/8| < a million b. - 8 < (x-8) < 8 ; subsequently for the sequence based around a = 8, the radius of convergence R is 8 or |x-a|