Find the interval & radius of convergence for the power series in #1b?

Answer 1

interval: #1<=x<9#
radius: #4#

use the ratio test:

[note: for the series #sum_(n=1)^ooa_n#, if #lim_(nrarroo)abs(a_(n+1)/a_n)#
-is #<1#, series converges
-is #=1#, inconclusive
-is #>1#, series diverges]

if you apply the ratio test to this: #sum_(n=1)^oo(x-5)^n/(n4^n)#, evaluating #lim_(nrarroo)abs(((x-5)^(n+1)/((n+1)(4^(n+1))))/((x-5)^(n)/((n)(4^(n))))# will show what x-values make the series converge or diverge.

simplifying the limit: #=lim_(nrarroo)abs(((x-5)/((n+1)(4)))/(1/((n)))#
#=lim_(nrarroo)abs((n(x-5))/(4n+4))#

[note: here you can divide both the numerator and denominator by #n#, because for whatever #n# value is used, the value inside the absolute value signs will stay the same (dividing by #n/n# or 1)]

#=lim_(nrarroo)abs(((n(x-5))/n)/((4n)/n+4/n))#
#=abs((x-5)/(4+0))#
#=abs((x-5)/4)#

back to the ratio test, the series can only converge if #abs((x-5)/4)<1# or #abs((x-5)/4)=1#

Case 1: #abs(x-5)<4#
#-4<x-5<4#
#1<x<9# (the solution interval must include these values)

Case 2: #abs((x-5)/4)=1#
#x-5=-4, x-5=4#
#x=1,9#

if #x=1#, series becomes: #sum_(n=1)^oo(1-5)^n/(n4^n)#
#=sum_(n=1)^oo(-4)^n/(n4^n)#
#=sum_(n=1)^oo(-1)^n/(n)#
this is the alternating harmonic series, which converges by the alternating series test

if #x=9#, series becomes: #sum_(n=1)^oo(9-5)^n/(n4^n)#
#=sum_(n=1)^oo(4)^n/(n4^n)#
#=sum_(n=1)^oo1/n#
this is the harmonic series, which diverges. here is a proof

so include #x=1# in the interval, too: #1<=x<9#

radius of convergence is half the difference between the upper and lower values for the interval #=(9-1)/2=4#

and here is a video with a similar problem

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Answer 2

To find the interval and radius of convergence for the power series in problem 1b, you can use the Ratio Test. The Ratio Test states that for a power series ( \sum_{n=0}^{\infty} a_n(x - c)^n ), the series converges absolutely if the limit ( \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| ) exists and is less than 1.

Apply the Ratio Test to the given power series to find the interval and radius of convergence. This involves finding the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}(x - c)^{n+1}}{a_n(x - c)^n} \right| ]

Simplify the expression and solve for ( x ) to determine the interval of convergence. The radius of convergence is then the distance from the center ( c ) to the nearest point where the series converges.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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