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

Question

∑ (from n=1 to infinity) (x−3)^n/2n

Power Series Question:



I have been getting stuck on the interval of convergence. So far I have gotten my radius of convergence to be 1. When I check my endpoints I end up having (-1)^n/2n for x=2 and for x=4 I have (1)^n/2n. Thanks for the help this is much appreciated. sorry for any confusion. x-3 is being raised to the power n. I.e (x-3)^n

Answer 1

You radius of convergence is correct.

Interval can be found by using

|x-3| < 1

x -3 <1 or x-3 >-1

2 < x < 4

We still have to check endpoints.

At x = 4 series us 1/n which is divergent.

At x = 2 , series is (-1)^n/n which is convergent.

Hence, interval of convergence is:

[2,4)

Answer 2

So, I just double-checked your radius of convergence, and it's indeed correct. I had to do it in case there might have been some trouble from that since the radius of convergence helps find the interval of convergence.



so 2 < x < 4 is our current interval or (2, 4) for short. Let's test the endpoints. What you have to do is plug in the endpoints and then use any of the old series tests to determine convergence or divergence.



x = 2



∑ (from n=1 to infinity) (-1)^n/2n



Use Alternating Series Test.



b_n = 1/2n



b_(n+1) = 1/(2(n+1))



1) lim as n goes to infinity for b_n = 0

2) b_(n+1) < b_n, this series is convergent.



So x = 2 yields convergence.



x = 4



∑ (from n=1 to infinity) (1)^n/2n



But (1)^n is 1 for all n, so the series becomes



∑ (from n=1 to infinity) 1/2n



This is basically



1/2 times ∑ (from n=1 to infinity) 1/n



1/n is the harmonic series, which is divergent. So x = 4 yields divergence. The interval of convergence is [2, 4).



Hope this helps.

Answer 3

i think of it is the sequence Sigma(n=0 to infinity){x^n/3^(n + a million)}. (If no longer, mimic my technique with the properly suited formula for a_n.) With a_n = x^n/3^(n + a million), the Ratio attempt has us inspect the ratio o |a_n/a_(n+a million)| = |x^(n + a million)/3^(n + 2)*32^(n + a million)/x^n| = |x|/3. We take the decrease of that ratio as n -> countless, that's |x|/3, and the sequence converges honestly whilst it is < a million, or |x| < 3. as a result, the Radius of Convergence is 3. For the era of convergence, we would desire to attempt the convergence whilst |x| = 3, because of the fact the Ratio attempt fails whilst the decrease = a million, =r whilst |x|/3 = a million. whilst x = 3, the sequence is Sigma {3^n/3^(n + a million)} = Sigma{a million/3}. This diverges because of the fact the words do no longer mindset 0. whilst x = -3, the sequence will become Sigma{(-a million^n)/3} and this diverges, lower back because of the fact the words do no longer mindset 0. hence, the era of Convergence is (-3,3).