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

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

Answer 1

To determine the values of (x) for which a series converges, we typically need to examine the convergence of the series. Without knowing the specific series provided in the question, it's challenging to give a precise answer.

However, some common tests for convergence include the ratio test, the root test, the comparison test, and the integral test. Each of these tests can help determine the convergence behavior of a series for a given range of values of (x).

If you provide the specific series, I can offer more guidance on which convergence test to apply and how to determine the values of (x) for which the series converges.

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Answer 2

#1<x<2#

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|

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|
#|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
#-1/2+3/2
#1
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
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Answer 3
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.

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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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