How do you find the radius of convergence #Sigma 1/(n!)x^(n^2)# from #n=[1,oo)#?

Answer 1

The series:

#sum_(n=1)^oo x^(n^2)/(n!)#

has radius of convergence #R=1#

We can us the ratio test:

#abs (a_(n+1)/a_n) = abs (((x^((n+1)^2))/((n+1)!))/((x^(n^2))/(n!))) = abs(x)^(n^2+2n+1)/abs(x)^(n^2) (n!)/((n+1)!) = abs(x)^(2n+1)/(n+1)#

Clearly we have that:

#L= lim_(n->oo) abs (a_(n+1)/a_n) =lim_(n->oo) abs(x)^(2n+1)/(n+1) = {(0" for " abs(x) <=1),(oo" for " abs(x) >1):}#
So the radius of convergence is #R=1#
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Answer 2

To find the radius of convergence of the series Σ(1/(n!)x^(n^2)) from n=1 to infinity, we can use the ratio test.

First, we apply the ratio test:

lim (|a_{n+1}/a_n|) as n approaches infinity.

Here, a_n represents the nth term of the series, which is 1/(n!)x^(n^2).

So, a_{n+1} = 1/((n+1)!)x^((n+1)^2).

Now, taking the ratio:

|a_{n+1}/a_n| = |((1/((n+1)!)x^((n+1)^2)) / (1/(n!)x^(n^2)))|.

Simplify the ratio:

|a_{n+1}/a_n| = |x^((n+1)^2-n^2) / ((n+1)!) / (n!)|.

Simplify the exponent and factorial terms:

|a_{n+1}/a_n| = |x^(2n+1) / (n+1)|.

Now, taking the limit as n approaches infinity:

lim |x^(2n+1) / (n+1)| as n approaches infinity.

For the series to converge, this limit must be less than 1. So, we consider:

|x^(2n+1) / (n+1)| < 1.

For x^(2n+1) / (n+1) to be less than 1, the absolute value of x^(2n+1) must be less than the absolute value of (n+1).

Hence, we have:

|x^(2n+1)| < |n+1|.

This holds true for all n, but we're interested in the radius of convergence, so we consider the limit as n approaches infinity:

lim |x^(2n+1)| / |n+1| as n approaches infinity.

Since x is a constant, and as n approaches infinity, x^(2n+1) will tend to either positive or negative infinity depending on the sign of x. Therefore, for the limit to be finite (and thus less than 1), x^(2n+1) must approach 0 as n approaches infinity.

Hence, the radius of convergence is 0, implying that the series converges only at x = 0.

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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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