Uniform and absolute convergence of complex series to $\log(1+z)$

Question

Show that the series $\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}\frac{z^n}{n}$ converges uniformly and absolutely to $\log(1+z)$ on the open disk where $\log(\rho e^{i\theta})=\log(\rho)+i\theta$ with $-\pi<\theta<\pi$.

I have no idea where to start.

Log of 1+z, not 1-z. // "Uniformly" even around z=-1, sure? // "I have no idea where to start." How come? — Did, Dec 29 '13 at 11:12

Haha · Answer 1 · 2013-12-29T16:25:25.930

1

It converges uniformly for $|z|\leq r<1$. Now we have that $$\frac {1}{z+1}=\sum_{n=0}^{\infty}(-z)^n=\sum_{n=0}^{\infty}(-1)^nz^n$$ (geometric series).

Let $$f(z)=\sum_{n=0}^{\infty}(-1)^n\frac {z^{n+1}}{n+1}$$, $|z|<1$.

Then $f:D(0,1)\to \Bbb C$ is infinitely differentiable and $$f'(z)=\frac {1}{z+1}$$.

Also we have that $log'(1+z)=\frac {1}{z+1}$ for $|z|<1$. and thus $log'(z+1)=f'(z)$ for every $|z|<1$. Now,integrate and find something more...

edited Dec 29 '13 at 16:25

answered Dec 29 '13 at 11:38

Haha

5,648

It converges uniformy on every compact set within its radius of convrrgence. It should be $|z|\leq r < 1$. – Mhenni Benghorbal Dec 29 '13 at 14:48
@MhenniBenghorbal correct.i forgot the $=$ symbol. – Haha Dec 29 '13 at 16:25

Uniform and absolute convergence of complex series to $\log(1+z)$

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