Region of convergence of the given series.

Question

I'm trying to find the region of convergence for the series $$\sum_{n=0}^{\infty}\frac{(n+1)z^n}{(1+z)^n}$$ and further show that the series is in fact a polynomial within that region.

By ratio test, we have $$\begin{align}\lim_{n\rightarrow\infty}\left|\frac{(n+2)z^{n+1}}{(1+z)^{n+1}}\cdot\frac{(1+z)^{n}}{(n+1)z^{n}}\right|&=\lim_{n\rightarrow\infty}\left|\frac{(n+2)z}{(n+1)(1+z)}\right|\\&=\left|\frac{z}{1+z}\right|\lim_{n\rightarrow\infty}\left|\frac{n+2}{n+1}\right|\\&=\left|\frac{z}{1+z}\right|\\&<1\end{align}$$

So the series converges for all $z$ such that $\left|\frac{z}{1+z}\right|$. But my question is, is this sufficient for the region of convergence? I tried to make it into a form of $|z-z_0|<R$, but it wasn't working.

Also, if this is the region, how do I know that the given series is polynomial? I understand that it is polynomial if $\frac{z}{1+z}$ is real. But I'm having trouble showing that.

Answer 1

$\left|\frac{z}{1+z}\right| < 1$ holds if $z$ is closer to $0$ than to $-1$, that is the half plane $\{ \operatorname{Re} >- \frac 12 \}$.

To compute the value of the series, consider $$ \sum_{n=0}^\infty (n+1) w^n $$ for $|w| < 1$, which is the derivative of $$ \sum_{n=0}^\infty w^{n+1} = w \sum_{n=0}^\infty w^{n} = \frac{w}{1-w} \, , $$ then substitute $w = \frac{z}{1+z}$.

Answer 2

Hints: The ratio test tells you the series converges absolutely for $|z/(1+z)|<1.$ This says $|z|<|1+z|$ or $|z-0| = |z-(-1)|.$ In other words, $z$ is closer to $0$ than it is to $-1.$ Continue thinking in this way to finish the convergence/divergence question.

For the polynomial conclusion, notice that $\sum_{n=0}^{\infty}(n+1)w^n$ is the derivative of $1/(1-w),$ which equals $1/(1-w)^2.$