Show that $\sum_{n=0}^\infty n^nx^n$ diverges for $x \neq 0$

Question

Show that the series $$\sum_{n=0}^\infty n^nx^n$$ diverges for $x \neq 0$.

Any help? I don't know where to start.

Answer 1

You have a power series whose radius of convergence $R$ is $0$:

So the root test is certainly appropriate:

$$\frac 1{\lim_{n \to \infty} \sqrt[\large n]{n^n}} = \frac {1}{\lim_{n\to \infty} n} = 0$$

Since the radius of convergence is $0$, the series diverges for any $x\,$ such that $ \;|x| > 0$.
I.e., the series diverges $\forall x \neq 0$.

Answer 2

Fix $x\neq 0$. Let $u_n=n^nx^n$. Show that $u_n\not\to 0$.

Answer 3

Use the root test to find that the radius of convergence of the power series is $0$--and so it converges only at its center point (at $0$).

Answer 4

For $x>0$, $\sum_{n=0}^{\infty} (xn)^n \geq \sum_{n=\lceil \frac{1}{x} \rceil}^{\infty} (xn)^n \geq \sum_{n=\lceil \frac{1}{x} \rceil}^{\infty} (x\lceil \frac{1}{x} \rceil)^n \geq \sum_{n=\lceil \frac{1}{x} \rceil}^{\infty} (1)^n$.