0
$\begingroup$

determine whether the series convergence

$\sum _{n=1}^{\infty \:}\frac{i^n}{n} $

My teacher said it is convergent but the ratio test is inconclusive and the root test is inconclusive

$\endgroup$
3
  • $\begingroup$ Does the sequence of partial sums converge? You may have to look at a few -- say, 4 -- subsequences. If they all converge, and to the same limit, then... $\endgroup$
    – Clement C.
    Nov 7 '16 at 21:31
  • 1
    $\begingroup$ @ClementC. $4$? Isn't $2$ better? $\endgroup$
    – b00n heT
    Nov 7 '16 at 21:32
  • $\begingroup$ Whatever works, I'd have gone for 4 (out of safety). $\endgroup$
    – Clement C.
    Nov 7 '16 at 21:33
1
$\begingroup$

The series is convergent but not absolutely convergent. Absolute convergence would say that the series $$ \sum_1^\infty \left| \frac{i^n}{n} \right| = \sum_1^\infty \frac{1}{n} $$ converges, and we know that is not the case.

However, we can break the series in question up as $$ \sum_{n=1}^\infty \frac{i^n}{n} = \sum_{m=1}^\infty (-1)^m \frac{1}{2m} + i \sum_{m=0}^\infty (-1)^m \frac{1}{2m+1} $$ and each of those alternating sign series can be shown to converge by grouping two terms together, getting a sum like $$ \sum_{m=1}^\infty (-1)^m \frac{1}{2m} = \frac12 \sum_{k=1}^\infty \left(\frac{1}{k} - \frac{1}{k+1} \right) = \frac12 \sum_{k=1}^\infty \frac1{k^2+k} $$ which converges by the ratio test.

$\endgroup$
1
  • $\begingroup$ The last lines are unnecessary if you quote Leibniz Series Theorem. +1 $\endgroup$
    – DonAntonio
    Nov 7 '16 at 21:55
0
$\begingroup$

Hint: Subdivide the Series in two sub-series: one over $2n$ and one over $2n+1$ and show that these converge.

$\endgroup$

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.