What is the interval of convergence of #sum_1^oo [(2n)!x^n] / ((n^2)! )#?

Answer 1

The series:

#sum_(n=0)^oo ( (2n)!x^n)/((n^2)!)#

is absolutely convergent for #x in (-oo,+oo)#

The ratio test can be used by assessing:

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

Clearly:

#lim_(n->oo) abs (a_(n+1)/a_n) = abs(x)*lim_(n->oo) ((2n+2)(2n+1))/ ((n^2+2n+1)(n^2+2n)* ... * (n^2+1)) = 0 #
for every #x#, thus the series is convergent for #x in RR#.
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Answer 2

The interval of convergence for the series ( \sum_{n=1}^{\infty} \frac{(2n)!x^n}{(n^2)!} ) can be determined using the ratio test. By applying the ratio test, we examine the limit:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(2(n+1))!x^{n+1}}{((n+1)^2)!} \cdot \frac{(n^2)!}{(2n)!x^n} \right| ]

After simplifying and taking the limit, we find that the ratio is ( |x|/4 ). For the series to converge, this ratio must be less than 1, which implies that ( |x| < 4 ).

Therefore, the interval of convergence is ( -4 < x < 4 ).

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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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