For which $z$ does $\sum_{n=1}^\infty \left[\frac {z(z+n)}{n}\right]^{n}$ converge? Here is one of my olympiad problem, I have to find the radius of following series
$$\sum_{n=1}^\infty \left[\frac {z(z+n)}{n}\right]^{n}$$
And here is my attempt
$$U_{n}= \left[\frac {z(z+n)}{n}\right]^{n}$$
$$U_{n+1}=\left[\frac {z(z+n+1)}{n+1}\right]^{n+1}$$
but the problem is when I take $\lim_{n\to \infty} \left|\frac {U_{n+1}}{U_{n}}\right|$ 
But it will leave me ugy terms which i cant see any good way to see the center or radius of circle , could you give me some hint ?
 A: Although, as has been pointed out, this is not a power series, it is a series of complex numbers $a_n=\left[\frac {z(z+n)}{n}\right]^{n}$ so one can apply the ratio test (or other convergence tests for series) to determine whether there are conditions under which the series converges.
\begin{eqnarray}\frac{\left[\frac {z(z+n+1)}{n+1}\right]^{n+1}}{ \left[\frac {z(z+n)}{n}\right]^{n}}
&=&z\left(\frac{z+n+1}{z+n}\right)^n\cdot\frac{1}{\left(1+\frac{1}{n}\right)^n}\cdot\left(\frac{z+n+1}{n+1}\right)\\
&=&z\left(1+\frac{1}{z+n}\right)^n\cdot\frac{1}{\left(1+\frac{1}{n}\right)^n}\cdot\left(1+\frac{1}{z+n}\right)\\
&=&z\left(\frac{1+\frac{1}{z+n}}{1+\frac{1}{n}}\right)^n\cdot\left(1+\frac{1}{z+n}\right)
\end{eqnarray}
\begin{equation}
\lim_{n\to\infty}\left\vert z\left(\frac{1+\frac{1}{z+n}}{1+\frac{1}{n}}\right)^n\cdot\left(1+\frac{1}{z+n}\right)\right\vert=\vert z\vert<1\end{equation}
Therefore the (non-power) series will converge for $\vert z\vert<1$ if that is what one wishes to call the "radius" of the series.
A: Let $z\in \mathbb C.$ Then $((z+n)/n)^n = (1+z/n)^n \to e^z\ne 0.$ Hence for large enough $n,$ $0<(1/2)|e^z| < |(z+n)/n)^n| < 2|e^z|.$ For such $n$ we then have
$$(1/2)|e^z||z|^n \le \left | \frac{z(z+n)}{n}\right|^n \le 2|e^z||z|^n.$$
Thus the terms of our series do not converge to $0$ if $|z|\ge 1,$ and they are bounded in absolute value by a constant times $|z|^n$ if $|z|<1.$ Thus the series converges iff $|z|<1.$
A: (1). For $|z|<1$ let $|z|=1-r$ with $1>r>0.$ For all sufficiently large n we have $|z^2/n|<r/2$ and $$|z(z+n)/n|=|z+z^2/n|\leq |z|+|z^2/n|<(1-r)+r/2=(1-r/2).$$ So the series converges by comparison with the geometric series $\sum_n(1-r/2)^n.$
(2). For $|z|>1$ let $|z|=1+r$ with $r>0.$ For all sufficiently large $n$ we have $|z^2/n|<r/2$ and $$|z(z+n)/n|=|z+z^2/n|\geq |z|-|z^2/n|>(1+r)-r/2>1.$$  So the terms of the series do not $\to 0$ and the series diverges.
