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Does this series converge or diverge? $$\sum_{k=1}^{\infty}\ln(\frac{k}{k+1})$$

my thought is that, I can break it down to $$\sum_{k=1}^{\infty}\ln(k) - \sum_{k=1}^{\infty}\ln(k+1)$$ then maybe using comparison test or something? But I don't know exactly how to prove whether this series converges or diverges.

Any help would be appreciated!

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What you broke down is wrong, because $\sum \ln k $ diverges. Instead, you can calculate the partial sum directly. Let $$ S_n=\sum_{k=1}^n \ln (k/(k+1)), $$ then it is equal to $$ S_n=-\ln(n+1). $$ Therefore, given series diverges.

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You're close to the answer. Just study the $S_n$ which is the partial sum up to $n$. Infact, $S_n = \displaystyle \sum_{k=1}^n \left(\ln k - \ln(k+1)\right) = - \ln(n+1) \to -\infty$, thus the series diveges to $-\infty$.

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    $\begingroup$ @Ramiro Considering that $\ln(k/(k+1) = \ln(k) - \ln(k+1)$, this answer is right on target, and your comment/review is out of line. You might consider deleting it. $\endgroup$ – Bungo Apr 18 '16 at 2:36
  • $\begingroup$ @Kf-sansoo +1 for the concise and clear post! -Mark $\endgroup$ – Mark Viola Apr 18 '16 at 3:36
  • $\begingroup$ @Bungo Kf-Sansoo edited his answer AFTER my review. When I reviewed his answer (for low quality) it was only one line: "You're close to the answer. Just study the Sn which is the partial sum up to n". Such kind of one line hint is not really an answer and it should be presented as a comment. $\endgroup$ – Ramiro Apr 18 '16 at 4:01
  • $\begingroup$ @Dr.MV Kf-Sansoo edited his answer AFTER my review. When I reviewed his answer (for low quality) it was only one line: "You're close to the answer. Just study the Sn which is the partial sum up to n". Such kind of one line hint is not really an answer and it should be presented as a comment. $\endgroup$ – Ramiro Apr 18 '16 at 4:02

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