Consider the series $$\sum{\frac{i^{n}}{n}}$$

We know that this series is not absolutely convergent as $\sum{|z_{n}|}$ gives harmonic series which is divergent.

However this series could be conditionally convergent or divergent.

I applied Ratio test (and got $|\frac{z_{n + 1}}{z_{n}}|$ approaches $1$) and Root test (which gave $\root{n}\of{|z_{n}|} = {(\frac{1}{n})}^{\frac{1}{n}}$ which is an indeterminate form as n approaches infinity. On manually checking for large values of n this approaches 1 though) but I am unable to certainly find an answer.

Another observation is that Root Test and Ratio test take mod and hence the result of these tests will not be different from what we will obtain for harmonic series. How will we then check if a series is conditionally convergent or not using these tests. Are these tests only to check absolute convergence?

  • 2
    $\begingroup$ Try separating out the real and imaginary parts. $\endgroup$
    – saulspatz
    Nov 16, 2019 at 16:56
  • 1
    $\begingroup$ Of course it is convergente, you should notice that $i^{4p+1}=i$, $i^{4p+2}=-1$, $i^{4p+2}=-i$ and $i^{4p+3}=-i$ , $i^{4p+4}=1$. By spletting your sum into two parts, your serie converges since $\sum_{i=1}^\infty (-1)^n/n$ converges. $\endgroup$
    – Gustave
    Nov 16, 2019 at 16:58

3 Answers 3


The series is convergent by Dirichlet's Test, since $\;\sum\limits_{n=1}^\infty i^n\;$ is a bounded series and $\;\left\{\frac1n\right\}\;$ is a decreasing sequence convergent to zero.


Take a look at this graphical representation of your series :

enter image description here

which enters into the (disputed!) category of "graphical proofs". (Therefore, I do not pretend that it is a full fledge proof !)

This inward spiraling is a testimony of the convergence of the series.

But to which complex number ? To

$$-\frac12 \log(2)+i \frac{\pi}{4} \approx -0.3466+0.7854 i \tag{1}$$

(check it on the figure !). Why that ? Because your expression, now that we are confident that it is a convergent sequence, is the value in $x=i$ of the classical series :


Converting $1-i$ into its equivalent trigonometric form $\sqrt{2}e^{-i \tfrac{\pi}{4}}$, (2) is transformed into

$$-\ln(1-i)=-\log(\sqrt{2}e^{-i \tfrac{\pi}{4}})\tag{3}$$

finally giving expression (1)

Important remark : in fact, in (3), we have manipulated the complex logarithm function as the familiar real logarithm function. We are fortunate that we are in a case where it is harmless. But if one day you have lectures on complex function theory, you will be introduced to all subtleties of $\log:\mathbb{C}\to \mathbb{C} $ and its "branches" (we have been using the "principal branch" here)...




For definiteness assume $n=2m+1$;

$\Re S_{n}= \sum_{k=1}^{m} (-1)^k/(2k)$;

$\Im S_n= \sum_{k=0}^{m}(i^{2k+1})/(2k+1)=i(-1)^k/(2k+1)$;

$S_n$ converges $\iff \Re S_n$ and $\Im S_n$ converge.

Leibniz criterion : $\Re S_n$, $\Im S_n$ converge.

Cf. comments by Saulspatz and Gustave.


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.