How do you find the radius of convergence #Sigma x^(6n)/(n!)# from #n=[1,oo)#?

Answer 1

The answer is #-oo < x < +oo #

Apply the ratio test to compute the convergence interval

#lim_(n->oo)|((x^(6(n+1))/((n+1)!))/(x^(6n)/(n!)))|#
#=lim_(n->oo)(|x^6/(n+1)|)#
#=|x^6|lim_(n->oo)(|1/(n+1)|)#
#=|x^6|*0#
#=0#
As the limit is #<1# for every #x#, therefore #sum_(n=1)^(+oo)x^(6n)/(n!)# converges for all #x#
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Answer 2
To find the radius of convergence for the series \( \sum_{n=1}^\infty \frac{x^{6n}}{n!} \), we can use the ratio test. The ratio test states that if \( L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| \), then the series converges absolutely if \( L < 1 \), diverges if \( L > 1 \), and the test is inconclusive if \( L = 1 \). In our case, \( a_n = \frac{x^{6n}}{n!} \), so: \[ \frac{a_{n+1}}{a_n} = \frac{x^{6(n+1)}}{(n+1)!} \cdot \frac{n!}{x^{6n}} = \frac{x^6}{n+1} \] Taking the limit as \( n \) approaches infinity: \[ L = \lim_{n \to \infty} \left| \frac{x^6}{n+1} \right| = 0 \] Since this limit is always less than 1 for any real value of \( x \), the series converges absolutely for all real numbers \( x \). Therefore, the radius of convergence is \( R = \infty \).
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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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