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Question

# Find the series’ radius and interval of convergence. For what values of x does the series converge?$\sum _ { n = 1 } ^ { \infty } n ^ { n } x ^ { n }$

Solution

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Let the given series be $\displaystyle\sum_{n=1}^{\infty} u_n = \displaystyle\sum_{n=1}^{\infty} n^nx^n$. Apply the Ratio Test to the series $\displaystyle\sum_{n=0}^{\infty} |u_n|$, where $u_n$ is the $nth$ term of the given power series. Now \begin{align*} \displaystyle \lim_{n \to \infty} \Big|\frac{u_{n+1}}{u_n} \Big| = & \displaystyle \lim_{n \to \infty} \Big| \frac{(n+1)^{n+1}x^{n+1}}{ n^nx^n} \Big| \\ = & \displaystyle \lim_{n \to \infty} \Big| \frac{(n+1)^{n+1}x}{n^n} \Big| \\ = & |x|\displaystyle \lim_{n \to \infty} \Big( \frac{(n+1)^{n}(n+1)}{n^n} \Big) \\ = & |x|\displaystyle \lim_{n \to \infty} \Big( \Big(\frac{n+1}{n}\Big)^{n}(n+1) \Big) \\ = & |x|\Big(\displaystyle \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n \Big) \Big( \displaystyle\lim_{n \to \infty} (n+1) \Big) \\ = & e|x| \displaystyle\lim_{n \to \infty} (n+1) ,\quad \text{as } \lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n=e \end{align*} Notice that if $x\neq 0$ then $e|x| \displaystyle\lim_{n \to \infty} (n+1) =\infty$. Also if $x=0$ then $|x| (e) \displaystyle\lim_{n \to \infty} (n+1) =0< 1$. Therefore the given series is absolutely convergence only if $x=0$.

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