How do you find the radius of convergence #Sigma n^nx^n# from #n=[1,oo)#?

#sum_(n=1)^oo n^nx^n#

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

#lim_(n->oo) abs (a_(n+1)/a_n) = lim_(n->oo) (n+1)(1+1/n)^nabs(x)#

#lim_(n->oo) (1+1/n)^n = e#

for any #x>0# we have:

#lim_(n->oo) abs (a_(n+1)/a_n) = oo #

and the series is divergent, so the radius of convergence is #R=0#

To find the radius of convergence of the series \(\sum_{n=1}^{\infty} n^n x^n\), you use the ratio test. The ratio test states that if \(L\) is the limit of the absolute value of the ratio of consecutive terms, then the series converges if \(L < 1\) and diverges if \(L > 1\). The radius of convergence \(R\) is the reciprocal of the limit \(L\), if \(L\) exists. To apply the ratio test, you find the limit of \(\left|\frac{a_{n+1}}{a_n}\right|\) as \(n\) approaches infinity, where \(a_n = n^n x^n\). Taking the ratio of consecutive terms, we have: \[ \lim_{n \to \infty} \left|\frac{(n+1)^{n+1}x^{n+1}}{n^n x^n}\right| = \lim_{n \to \infty} \left|\frac{(n+1)^{n+1}}{n^n}\right| \cdot |x| \] Using the fact that \(\lim_{n \to \infty} \frac{(n+1)^{n+1}}{n^n} = e\), the limit simplifies to \(e|x|\). For convergence, \(e|x| < 1\). Thus, the radius of convergence is \(R = \frac{1}{e}\).