Sum the series $\sum_{n=1}^\infty \frac{z^n}{n}$

Asked Mar 21 '24 at 16:51

Active Mar 21 '24 at 18:04

Viewed 66 times

The problem asks to sum it in its circle of convergence.

It's radius of convergence is $R=lim_{n \to \infty} \left|{\frac{\frac{1}{2n+1}}{\frac{1}{2n+2}}}\right|$=1. And when $\left|z\right|=1$ it diverges by the Limit Comparison Test with $1/n$. So its circle of convergence is $\left|z\right|<1$, id est, $z \in D(0,1)$.

We have that $$ \left(\sum_{n=1}^\infty \frac{z^n}{n} \right)' = \sum_{n=1}^\infty z^{n-1} = -1+\sum_{n=0}^\infty z^{n} = -1 + \frac{1}{1-z} $$ when $\left|z\right|<1$, so doing the integration we have $$ \sum_{n=1}^\infty \frac{z^n}{n} = \int \left(-1 + \frac{1}{1-z}\right)dz=-z-\log(1-z)+C $$ But the thing is, ¿what is the value of that C?

edited Mar 21 '24 at 18:04

asked Mar 21 '24 at 16:51

baristocrona

3

Hint: Take $z=0$ on both sides. – Liding Yao Mar 21 '24 at 16:52
Thank you Linding, now I see. – baristocrona Mar 21 '24 at 16:54
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There is a mistake in the first line of your derivation. The first term ($-1$) should not be there. – Fabian Mar 21 '24 at 17:58
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@Fabian why not? isn't it necessary to supply the $z^0=1$ that we add to the series? – baristocrona Mar 21 '24 at 18:08
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@baristocrona you are not adding $1$ to the series. You are only reindexing it. $\sum_{n=1}^\infty z^{n-1}=\sum_{m=0}^\infty z^m$, by the substitution $m=n-1$. – Anne Bauval Mar 21 '24 at 18:23
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Does this answer your question? Closed form of series $z^n/n$. Or some other post of this list? – Anne Bauval Mar 21 '24 at 18:26
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@AnneBauval thanks for the explanation. – baristocrona Mar 21 '24 at 18:32

Sum the series $\sum_{n=1}^\infty \frac{z^n}{n}$

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