Find the radius of convergence and interval of convergence of the power series?

Question

This question is new to me, if you could please help me with it with full explanation

Find the radius of convergence and interval of convergence of the power series
∞∑ n=0 n (x^n) / 3^(n+1)


Thanks

Answer 1

I think this is the series Sigma(n=0 to infinity){x^n/3^(n + 1)}. (If not, mimic my procedure with the correct formula for a_n.) With a_n = x^n/3^(n + 1), the Ratio Test has us look at the ratio o |a_n/a_(n+1)| = |x^(n + 1)/3^(n + 2)*32^(n + 1)/x^n| = |x|/3. We take the limit of that ratio as n -> infinite, which is |x|/3, and the series converges absolutely when this is < 1, or |x| < 3. Thus, the Radius of Convergence is 3.



For the interval of convergence, we must test the convergence when |x| = 3, since the Ratio Test fails when the limit = 1, =r when |x|/3 = 1. When x = 3, the series is Sigma {3^n/3^(n + 1)} =

Sigma{1/3}. This diverges because the terms do not approach 0.



When x = -3, the series becomes Sigma{(-1^n)/3} and this diverges, again because the terms do not approach 0.



Therefore, the Interval of Convergence is (-3,3).

Answer 2

What bit is new? That you do not understand what it is asking? You need to find the values of x for which this converges.



You can rewrite it as:



(1/3) ∞∑ n=0 n (x/3)^n



Now it should be obvious from that form that the series diverges if |x/3| >= 1



And it can be shown to converges for |x/3| < 1. Replacing x/3 with r and omitting the 1/3:

∞∑ n=0 n r^n

= ∞∑ n=1 n r^n

= ∞∑ n=1 r^n + ∞∑ n=2 r^n + ∞∑ n=3 r^n ...

= (1 + r + r^2 + ...)( ∞∑ n=1 r^n)

= ( ∞∑ n=0 r^n) ( ∞∑ n=1 r^n)

... which is the product of two geometric series, which is finite if |r| < 1

Answer 3

to seek out < R, use the ratio or root attempt (in many situations, ratio attempt will yield the resultant period). employing the ratio experiment ]. 2. Simplify the expression [(x-8)^(n+a million) / 8^(n+a million)] * [8^n / (x-8)^(n)] interior the reduce: [(x-8)^n * (x-8)^a million * 8^n] / [(x-8)^n * 8^n * 8^a million]; (x-8)^n time period and eight^n time period cancel leaving: (x-8)/8. 3. The reduce simplifies to lim_n-inf._[ because the expression does no longer position self belief in 'n', that you would possibly want to address it as a mild. employing the reduce-consistent rule, lim_n-inf._[ (x-8)/88)/8. 4. in preserving with the ratio attempt for the sum of a set ) < a million, then the sequence converges for sure (If the sequence converges actual, there's a theorem that asserts that the sequence also) > a million or equals infinity or DNE, then the sequence diverges. hence, the sequence with the sequence (x-8)^n / 8(x-8)/eightotherwise. 5. to locate the radius of convergence by employing technique of manipulating the inequality you ended with after taking the reduce which hence is (x-8)/8(x-8)/8eight < (x-8) < 8 ; therefore for the sequence based round a = 8, the radius of convergence R is 8