Radius convergence of power series How would I go about finding the radius of convergence using a limit ratio test? Can I get a hint for this one?
$\displaystyle\sum_{j=1}^\infty\frac{(jx)^j}{j!}$
 A: You can use ratio test. The $j$-term is equal to 
$$a_j=\frac{(jx)^j}{j!}.$$
Therefore, 
$$\left|\frac{a_{j+1}}{a_j}\right|=\left|\frac{\frac{[(j+1)x]^{j+1}}{(j+1)!}}{\frac{(jx)^j}{j!}}\right|=\frac{(j+1)^{j+1}|x|}{j^j}\cdot\frac{j!}{(j+1)!}
=\frac{(j+1)^j}{j^j}|x|=\left(1+\frac{1}{j}\right)^j|x|.$$
Therefore, we have 
$$\lim_{j\to\infty}\left|\frac{a_{j+1}}{a_j}\right|=\lim_{j\to\infty}\left(1+\frac{1}{j}\right)^j|x|=e|x|.$$
Therefore, $\lim_{j\to\infty}\left|\frac{a_{j+1}}{a_j}\right|<1$ if and only if $e|x|<1$, i.e. $|x|<\frac{1}{e}$. 
Now you also need to check whether the series converges at $x=\frac{1}{e}$ and at $x=-\frac{1}{e}$. I will leave these to you.
A: Hint:
$$\frac{(k+1)^{k+1}|x|^{k+1}}{(k+1)!}\frac{k!}{k^k|x|^k}=\left(1+\frac{1}{k}\right)^k|x|\xrightarrow[k\to\infty]{} e|x|\stackrel ?<1\ldots$$
A: Perhaps you might write the test ratio as
$$ | \frac{(j+1)^{j+1} \cdot x^{j+1} \cdot j!}{j^{j} \cdot x^{j} \cdot (j+1)!} | $$
$$ = | (j+1) \cdot ( 1 + \frac{1}{j} )^{j} \cdot \frac{1}{j+1} \cdot x |$$
$$ = | ( 1 + \frac{1}{j} )^{j} \cdot x |  .$$
Now you have something which is relatively simple to take the limit of as $j \rightarrow \infty$  .
A: Let $ a_n=\frac {n^n}{n!}$then $\frac {a_{n+1}}{ a_n  }=(1+\frac {1}{n})^n  $obviosly $ \lim\frac {a_{n+1}}{ a_n  }=e $
A: The first step is to factor $(jx)^j$ into $j^j$ and $x^j$. Then, use the ratio test. The ratio becomes $[(j+1)^(j+1)x]/(j^j)(j+1)$, $x \to \infty$, or $x*[(j+1)/(j)]^(j+1)$, $x \to\infty$. This diverges for all $x$.
