Evaluation of $\sum_{n=1}^{\infty} \frac {x^{-n}}{n}$

I wanted to evaluate

$$\sum_{n=1}^{\infty} \frac {2^{-n}}{n}$$

And noticed that for any base it has a pattern, so now I want to know how to evaluate

$$\sum_{n=1}^{\infty} \frac {x^{-n}}{n}$$

I don't have any approach. The result is logarithmic. If any problems occur $\forall x$ then I want a solution for valid $x$ values.

• Replace $x$ by $1/y$ and recognize the Taylor expansion of $-\log(1-y)$. – Claude Leibovici Sep 1 '14 at 10:41
• You can try differentiating term by term, summing the resultant series and integrating the resulting formula (being careful about the constant of integration) – Mark Bennet Sep 1 '14 at 10:45

One alternative way, that generalizes to a wide range of similar questions, is to differentiate with respect to $x$ under the sum: $$f(x)=\sum_{n=1}^{\infty} \frac {x^{-n}}{n}\\ f'(x)=-\sum_{n=1}^{\infty}x^{-n-1}$$ which is just a geometric series, which can be evaluated, and subsequently integrated to retrieve $f(x)$. You'll need to fix the constant of integration, which can be set by what happens as $x\to\infty$.

Substitute the $y = -1/x$ and you'll notice than this is a Taylor series for logarithm ($-\ln(1+y)$):

Also: $$\ln(1+y) = y - \frac{y^2}{2} + \frac{y^3}{3} - \cdots$$

• @monhawk - very clever solution. a minor typo but you probably meant to say $\ln(1+y)$. – hypergeometric Sep 1 '14 at 14:41

We want to compute the following sum: $$\sum_{n=1}^{+\infty}\frac{1}{nz^n},\;\;\;\; z\in\mathbb C\;.$$ We immediately see that $|z|>1$, in order to have absolute convergence.

We recall first two results:

$\bullet\;\;$First: $$\log(1+z)=\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{z^n}{n},\;\;\;\forall |z|<1$$ $\bullet\;\;$Second: $$\prod_{n=0}^{+\infty}\left(1+z^{2^{n}}\right)= \sum_{n=0}^{+\infty}z^{n}=\frac{1}{1-z},\;\;\;\forall |z|<1$$ The last one can be proved, showing by induction that $\prod_{k=0}^{N}\left(1+z^{2^{k}}\right)=\sum_{k=0}^{2^{N+1}-1}z^{k}$.

Ok: \begin{align*} \sum_{n=1}^{+\infty}\frac{1}{nz^n}=& \sum_{n=1}^{+\infty}\frac{1}{n}\left(\frac{1}{z}\right)^n\\ =&\underbrace{\sum_{k=0}^{+\infty}\frac{1}{2k+1}\left(\frac{1}{z}\right)^{2k+1}- \sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}}_{\log\left(1+\frac{1}{z}\right)}+ 2\sum_{k=1}^{+\infty}\frac{1}{2k}\left(\frac{1}{z}\right)^{2k}\\ =&\log\left(1+\frac{1}{z}\right)+ \sum_{k=1}^{+\infty}\frac{1}{k}\left(\frac{1}{z^2}\right)^{k}\\ =&\log\left(1+\frac{1}{z}\right)+ \log\left(1+\frac{1}{z^2}\right)+\cdots\\ =&\sum_{n=0}^{+\infty}\log\left(1+\frac{1}{z^{2^n}}\right)\\ =&\log\left(\prod_{n=0}^{+\infty}\left(1+\left(\frac{1}{z}\right)^{2^n}\right)\right)\\ =&\log\left(\frac{z}{z-1}\right) \end{align*}

If otherwise we want to sum

$$\sum_{n=1}^{+\infty}\frac{(-1)^n}{nz^n}$$ it's simpler: \begin{align*} \sum_{n=1}^{+\infty}\frac{(-1)^n}{nz^n}& =-\sum_{n=1}^{+\infty}(-1)^{n+1}\frac{1}{n}\left(\frac{1}{z}\right)^n\\ &=-\log\left(1+\frac{1}{z}\right)\\ &=\log\left(\frac{z}{z+1}\right). \end{align*}