Find the values of #x# for which the following series is convergent?

#\sum_(n=0)^(\infty)(2x-3)^n#

2 Answers
Apr 8, 2018

#1<x<2#

Explanation:

When trying to determine the radius and/or interval of convergence of power series such as these, it is best to use the Ratio Test, which tells us for a series #suma_n#, we let

#L=lim_(n->oo)|a_(n+1)/a_n|#.

If #L<1# the series is absolutely convergent (and hence convergent)

If #L>1#, the series diverges.

If #L=1,# the Ratio Test is inconclusive.

For Power Series, however, three cases are possible

a. The power series converges for all real numbers; its interval of convergence is #(-oo, oo)#
b. The power series converges for some number #x=a;# its radius of convergence is zero.
c. The most frequent case, the power series converges for #|x-a|<R# with an interval of convergence of #a-R<x<a+R# where we must test the endpoints to see what happens with them.

So, here,

#a_n=(2x-3)^n#

#a_(n+1)=(2x-3)^(n+1)=(2x-3)(2x-3)^n#

So, apply the Ratio Test:

#lim_(n->oo)|((cancel((2x-3)^n)(2x-3))/cancel((2x-3)^n))|#

#|2x-3|lim_(n->oo)1=|2x-3|#

So, if #|2x-3|<1#, the series converges. But we need this in the form #|x-a|<R:#

#|2(x-3/2)|<1#

#2|x-3/2|<1#

#|x-3/2|<1/2# results in convergence. The radius of convergence is #R=1/2.#

Now, let's determine the interval:

#-1/2<x-3/2<1/2#

#-1/2+3/2<x<1/2+3/2#

#1<x<2#

We need to plug #x=1, x=2# into the original series to see if we have convergence or divergence at these endpoints.

#x=1: sum_(n=0)^oo(2(1)-3)^n=sum_(n=0)^oo(-1)^n# diverges, the summand has no limit and certainly doesn't go to zero, it just alternates signs.

#x=2: sum_(n=0)^oo(4-3)^n=sum_(n=0)^oo1# diverges as well by the Divergence Test, #lim_(n->oo)a_n=lim_(n->oo)1=1 ne 0#

Therefore, the series converges for #1<x<2#

Apr 8, 2018

We can use the ratio test which says that if we have a series
#sum_(n=0)^ooa_n#

it is definitely convergent if:
#lim_(n->oo)|a_(n+1)/a_n|<1#

In our case, #a_n=(2x-3)^n#, so we check the limit:
#lim_(n->oo)|(2x-3)^(n+1)/(2x-3)^n|=lim_(n->oo)|((2x-3)cancel((2x-3)^n))/cancel((2x-3)^n)|=#

#=lim_(n->oo)|2x-3|=2x-3#

So, we need to check when #|2x-3|# is less than #1#:

I made a mistake here, but the above answer has the same method and a correct answer, so just have a look at that instead.