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

Answer 1

The interval of convergence is #x=0# and the radius of convergence is #R=0#.

#sum_(n=1)^oo(n!(x^n))/sqrt(n^n)=sum_(n=1)^oo(n!(x^n))/n^(n/2)#
The series #suma_n# is convergent when #L<1# when #L=lim_(nrarroo)abs(a_(n+1)/a_n)#. Here, this becomes:
#L=lim_(nrarroo)abs(((n+1)!(x^(n+1)))/(n+1)^(1/2(n+1))*n^(n/2)/(n!(x^n)))#

Simplifying:

#L=lim_(nrarroo)abs(((n+1)(n!))/(n!)*x^(n+1)/x^n*n^(n/2)/(n+1)^(n/2+1/2))#
#L=lim_(nrarroo)abs((n+1)/(n+1)^(n/2+1/2)*x*n^(n/2))#
The #x# can be moved to outside the limit, since the limit only depends on #n#.
Also note that #(n+1)/(n+1)^(n/2+1/2)=(n+1)^(1-(n/2+1/2))=(n+1)^(1/2-n/2)#.
#L=absxlim_(nrarroo)abs((n+1)^(1/2-n/2)(n^(n/2)))#
#L=absx(lim_(nrarroo)abs((n+1)(n/(n+1))^n))^(1/2)#
Note that #lim_(nrarroo)((n+1)/n)=e#, so #lim_(nrarroo)(n/(n+1))=1/e#. Thus, the overall limit approaches #oo# since the remaining portion of #n+1# is unbounded.
Since the limit approaches infinity, the only time when #L<1# will be when #x=0#, as this makes #L=0#.
Thus the interval of convergence is #x=0# and the radius of convergence is #R=0#.
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Answer 2
To find the radius of convergence of the series \(\sum_{n=1}^{\infty} \frac{n!x^n}{\sqrt{n^n}}\), we can use the ratio test. The ratio test states that if \(\lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = L\), then the series converges if \(L < 1\) and diverges if \(L > 1\). If \(L = 1\), the test is inconclusive. In this case, \(a_n = \frac{n!x^n}{\sqrt{n^n}}\), so: \[ \begin{align*} \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| &= \lim_{n \to \infty} \left| \frac{(n+1)!x^{n+1}}{\sqrt{(n+1)^{n+1}}} \cdot \frac{\sqrt{n^n}}{n!x^n} \right| \\ &= \lim_{n \to \infty} \left| \frac{(n+1)x}{\sqrt{n+1}} \right| \end{align*} \] Using L'Hôpital's rule, we evaluate the limit: \[ \begin{align*} \lim_{n \to \infty} \left| \frac{(n+1)x}{\sqrt{n+1}} \right| &= \lim_{n \to \infty} \left| \frac{x}{\frac{1}{2}(n+1)^{-\frac{1}{2}}} \right| \\ &= |2x| \end{align*} \] For the series to converge, \(|2x| < 1\). Thus, the radius of convergence is \(R = \frac{1}{2}\).
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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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