Consider $$h(z)=\sum_{n=1}^{\infty}\frac{(z-2)^n}{n}.$$ I wish to find an expression for $h$ as an elementary function.

This question has me stumped. I considered another function, $$f(z)=\sum_{n=1}^{\infty} n(z-2)^n.$$ This is much easier to express as an elementary function, as $$f(z)=\sum_{n=1}^{\infty} n(z-2)^n=(z-2)\frac{d}{dz}\sum_{n=1}^{\infty} (z-2)^n=\frac{z+3}{(z+2)^2}.$$ But for the function $h$, I cannot see a similar technique or a manipulation to yield such a function.

I would really appreciate a hint.


Hint: formally $h'(z)=\sum (z-2)^{n-1}$. Calculate this sum and integrate. The answer is $Log (3-z)$ for $|z-2| <1$ where $Log$ is the principle branch of logarithm. You have to have some knowledge of logarithms in the complex plane to answer this question.

  • $\begingroup$ Does this give $-\ln|3-z|-\ln|z-2|+C$? $\endgroup$ – JulianAngussmith Oct 22 '18 at 9:11
  • $\begingroup$ No, the sum of the series is not a real valued function. Please see my revised answer. $\endgroup$ – Kavi Rama Murthy Oct 22 '18 at 9:16
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    $\begingroup$ I believe I have calculated the sum incorrectly. I have that $$h(z)=\int\sum_{n=1}^{\infty} (z-2)^{n-1} \ dz=\int\sum_{n=0}^{\infty} (z-2)^n-\frac{1}{z-2} \ dz.$$ $\endgroup$ – JulianAngussmith Oct 22 '18 at 9:19

Taking the convergent part of this sum and making

$$ y = z-2 $$

we have

$$ h(y) = \sum_{k=1}^{\infty}\frac{y^k}{k} = \int_0^y\sum_{k=0}^{\infty}\zeta^kd\zeta = \int_0^y\frac{d\zeta}{1-\zeta} $$

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    $\begingroup$ Why does $$\sum_{k=1}^{\infty}\frac{y^k}{k}=\int\sum_{k=0}^{\infty} y^k\ dy?$$ $\endgroup$ – Bell Oct 22 '18 at 9:41
  • $\begingroup$ @Cesareo I don't understand how the second line of equations works. For instance, as $$\frac{y^k}{k} =\int y^{k-1} \ dy \ \ \text{doesn't this imply that} \ \ \sum_{k=1}^{\infty}\frac{y^k}{k}=\int\sum_{k=1}^{\infty}y^{k-1} \ dy?$$ $\endgroup$ – JulianAngussmith Oct 22 '18 at 23:00
  • $\begingroup$ @JulianAngussmith $y^0+y^1\Rightarrow\int_0^y\Rightarrow\frac{y^1}{1}+\frac{y^2}{2}$ $\endgroup$ – Cesareo Oct 22 '18 at 23:26
  • $\begingroup$ Is a definite integral necessary for this problem? $\endgroup$ – JulianAngussmith Oct 23 '18 at 0:29


Instead of looking at $h$, first try to calculate what $$\frac{\partial h}{\partial z}$$ would be.


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