complex series absolute convergence?

**chocaholic** · March 22nd 2011, 03:44 AM

Hi
I'm doing complex analysis and the section on convergence of series to infinity has me stumped

so any help would be great.

The question is as follows:
Show that the series:
sum of [z/(z+1)]^n from n=0 to infinity
converges absolutely if and only if R(z)>-1/2 where R(z) is the real part of z.

To me this looks like a power series problem with the need to find the radius of convergence but I don't know what to do with the denominator. I tried substituting [z/(z+1)] in the expansion of sum of (z)^n from n=0 to infinity=1/(1-z) and then finding the remainder term pn but this is firstly just for convergence (not absolute) and secondly I got stuck.I even tried using the ratio test but once again I got stuck. I know that for a series to be absolutely convergent the series of the absolute value of [z/(z+1)]^n must be convergent where IzI=sqrt(x^2+y^2) for z=x+iy, but I don't know how to incorporate all these seperate peices of information into a meaningful answer.

Any help would be greatly appreciated.

Thanks in advance.

**chisigma** · March 22nd 2011, 03:57 AM

Originally Posted by chocaholic

Hi
I'm doing complex analysis and the section on convergence of series to infinity has me stumped

so any help would be great.

The question is as follows:
Show that the series:
sum of [z/(z+1)]^n from n=0 to infinity
converges absolutely if and only if R(z)>-1/2 where R(z) is the real part of z.

To me this looks like a power series problem with the need to find the radius of convergence but I don't know what to do with the denominator. I tried substituting [z/(z+1)] in the expansion of sum of (z)^n from n=0 to infinity=1/(1-z) and then finding the remainder term pn but this is firstly just for convergence (not absolute) and secondly I got stuck.I even tried using the ratio test but once again I got stuck. I know that for a series to be absolutely convergent the series of the absolute value of [z/(z+1)]^n must be convergent where IzI=sqrt(x^2+y^2) for z=x+iy, but I don't know how to incorporate all these seperate peices of information into a meaningful answer.

Any help would be greatly appreciated.

Thanks in advance.

The series converges if...

$\displaystyle |\frac{z}{1+z}|<1 \implies |z|<|1+z|$ (1)

Now set in (1) $z=x + i\ y$ and find the values of x and y wchich satisfy (1)...

Kind regards

$\chi$ $\sigma$

**Prove It** · March 22nd 2011, 04:05 AM

You have the series $\displaystyle \sum_{n = 0}^{\infty}\left(\frac{z}{z + 1}\right)^n$ .

According to the ratio test, this series is convergent for all $\displaystyle z$ such that $\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| < 1$

$\displaystyle \lim_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right| = \lim_{n \to \infty}\left|\frac{\left(\frac{z}{z + 1}\right)^{n + 1}}{\left(\frac{z}{z + 1}\right)^n}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{\frac{z^{n+1}}{(z + 1)^{n + 1}}}{\frac{z^n}{(z + 1)^n}}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{z^{n+1}(z + 1)^n}{z^n(z + 1)^{n + 1}}\right|$

$\displaystyle = \lim_{n \to \infty}\left|\frac{z}{z + 1}\right|$

$\displaystyle = \left|\frac{z}{z + 1}\right|$ .

So the series is convergent for $\displaystyle \left|\frac{z}{z + 1}\right| < 1$

$\displaystyle \frac{|z|}{|z + 1|} < 1$

$\displaystyle |z| < |z + 1|$

$\displaystyle |z|^2 < |z + 1|^2$

$\displaystyle x^2 + y^2 < (x + 1)^2 + y^2$

$\displaystyle x^2 < (x + 1)^2$

$\displaystyle x^2 < x^2 + 2x + 1$

$\displaystyle 0 < 2x + 1$

$\displaystyle -1 < 2x$

$\displaystyle -\frac{1}{2} < x$ .

So the series is convergent for $\displaystyle z$ with $\displaystyle \textrm{Re}\,(z) > -\frac{1}{2}$

**chocaholic** · March 22nd 2011, 04:29 AM

Thank you so much. I feel so stupid I used the ratio test and got to lzl<lz+1l but it didn't occur to me to substitute z=x+iy in when it should have. Thank you so much you just helped me resolve 3 hours of work.

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