# How do you find the interval of convergence #Sigma (3^n(x-4)^(2n))/n^2# from #n=[1,oo)#?

And:

Test the intervals to see if the integral converges at the extremes:

Which converges as well. So the interval of convergence is:

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To find the interval of convergence for the series ∑(3^n(x-4)^(2n))/n^2 from n=1 to infinity, we will apply the ratio test.

First, we express the general term of the series:

a_n = (3^n(x-4)^(2n))/n^2

Next, we apply the ratio test:

lim (n→∞) |a_{n+1}/a_n|

= lim (n→∞) |[3^(n+1)(x-4)^(2(n+1))/((n+1)^2)] * [(n^2)/(3^n(x-4)^(2n))]|

= lim (n→∞) |(3(x-4)^2)/(n+1)^2|

To determine convergence, we take the limit:

lim (n→∞) |(3(x-4)^2)/(n+1)^2|

Since the limit depends on n, we evaluate it as n approaches infinity:

= |3(x-4)^2/∞|

= 0

Since the limit is less than 1, the series converges for all real numbers x.

Thus, the interval of convergence is (-∞, ∞).

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The interval of convergence for the given series is (|x - 4| < 3).

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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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