Calculus HW: Radius of Convergence of Geometric Series?

Question

I need to find the radius of convergence of the summation from 1 to infinity of (x^n)/(n*3^n). I tried the ratio test but it didn't work.

Answer 1

(x^n)/(n*3^n)



for x=0, f(0) = 0 + 0 + 0 + .... = 0



for | x | > 0, lim x>inf [(x^n)/(n*3^n)] = lim x>inf [(x^[n+1])/([n+1]*3^[n+1])



= lim x>inf [ (x)(x^n) / ( 3(3^n)(n+1) ]



= inf



so, since series only converges at x=0 and diverges everywhere else,

Radius of convergence = 0

Answer 2

Yes, this problem can be solved by using the ratio test.

Consider the ratio [(x^(n+1))/((n+1)*3^(n+1))] / [(x^n)/(n*3^n)].



The series converges when the limit (as n goes to infinity) of the absolute value of this ratio is less than 1, and diverges when this limit is greater than 1.



We really don't need to consider the case when this limit is 1, since we are asked only for the radius of convegence; this problem does not ask us to determine which endpoints are included or excluded. (However, if we were asked for the interval of convergence, we would have to consider this case and further analysis would be needed at each endpoint.)



lim n --> infinity |[(x^(n+1))/((n+1)*3^(n+1))] / [(x^n)/(n*3^n)]|

= lim n --> infinity |[nx^(n+1)*3^n] / [(n+1)x^n*3^(n+1)]|

= lim n --> infinity |(n/(n+1))x/3|

= |x/3| lim n --> infinity n/(n+1)

= |x/3| lim n --> infinity 1/(1 + 1/n)

= |x/3| / (1 + 0)

= |x/3|.



|x/3| < 1 when -3 < x < 3, so the series converges on (-3, 3).

|x/3| > 1 when x < -3 or x > 3, so the series diverges on (-infinity, -3) union (3, infinity).



So we know that the interval of convergence is (-3, 3), (-3, 3], [-3, 3), or [-3, 3]. Since the radius of convergence is the distance from the center of the interval of convergence to either of its endpoints(whether or not the endpoint is actually included in the interval), the radius of convergence is 3.



Lord bless you today!

Answer 3

to discover the era of convergence |x-a| < R, use the ratio or root attempt (often, ratio attempt will yield the consequent era). making use of 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 cut back: [(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 cut back simplifies to lim_n-inf._[ | (x-8)/8| ]. because of the fact the expression does not count on 'n', you could manage it as a relentless. utilising the cut back-consistent rule, lim_n-inf._[ (x-8)/8| ] = |(x-8)/8|. 4. in accordance to the ratio attempt for the sum of a series A_n from n = a million to infinity, if lim_n-inf._( |A_n+a million / A_n| ) < a million, then the series converges unquestionably (If the series converges unquestionably, there's a theorem that publicizes that the series additionally converges). If lim_n-inf._( |A_n+a million / A_n |) > a million or equals infinity or DNE, then the series diverges. hence, the series with the series (x-8)^n / 8^n converges if |(x-8)/8| < a million and diverges in any different case. 5. discover the radius of convergence via manipulating the inequality you ended with after taking the cut back which thus is (x-8)/8 < a million. a. |(x-8)/8| < a million b. - 8 < (x-8) < 8 ; hence for the series based around a = 8, the radius of convergence R is 8 or |x-a|