1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Replace $x$ by $1/y$ and recognize the Taylor expansion of $-\log(1-y)$. $\endgroup$ – Claude Leibovici Sep 1 '14 at 10:41
  • $\begingroup$ You can try differentiating term by term, summing the resultant series and integrating the resulting formula (being careful about the constant of integration) $\endgroup$ – Mark Bennet Sep 1 '14 at 10:45
7
$\begingroup$

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 $x\to\infty$.

$\endgroup$
5
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ @monhawk - very clever solution. a minor typo but you probably meant to say $\ln(1+y)$. $\endgroup$ – hypergeometric Sep 1 '14 at 14:41
4
$\begingroup$

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*}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.