# Convergence of $\sum_{n=1}^{\infty}(n+1)z^n$

Consider the series $\sum_{n=1}^{\infty}(n+1)z^n$

A) For which complex numbers $z$ does this series converge?

B) For those $z$, let $f(z)$ be the sum of the series and find $f(z)$.

C) Evaluate $f^{(5)}(0)$

My approach for A:

Using Ratio test, $$\lim_{n \to \infty}\frac{|a_n|}{|a_{n+1}|}=\lim_{n \to \infty}\frac{n+1}{n+2}=0$$

Therefore, radius of convergence is zero. Is this the right approach?

My approach for B

I'm guessing we have to use Taylor's Theorem $f(z)=\sum_{n=0}^{\infty} \frac{f^{(n)}(z_0)}{n!}(z-z_0)^n$

My Approach for C

I'm thinking I have to take the fifth derivative of something here.

I'd appreciate any help. Thank you

• The limit is $1$, not $0$. And you may recognize the given series as a derivative. Alternately, you can find an explicit formula for the partial sums, an bypass most theory. – André Nicolas May 1 '14 at 6:19
• The limit should be $1$. – Mhenni Benghorbal May 1 '14 at 6:19
• When $n$ is big, $\frac{n+1}{n+2}$ is real close to $1$. If you want to be formal, divide top and bottom by $n$, and let $n\to\infty$. – André Nicolas May 1 '14 at 6:22
• Note that $(n+1)/(n+2)=(1+1/n)/(1+2/n)$. Here both the numerator and the denominator tends to 1. – AD. May 1 '14 at 6:23
• The series for$\frac{1}{1-z}$ is familiar, it is $1+z+z^2+z^3+\cdots$, absolutely convergent if $|z|\lt 1$. The term by term derivative is $1+2z+3z^2+4z^3+\cdots$. This is $1+ \text{our given series}$. But the derivative of $\frac{1}{1-z}$ is $\frac{1}{(1-z)^2}$. – André Nicolas May 1 '14 at 14:12