I have following two series.

1.$\sum_{1}^{\infty}\ln\left(\frac{n}{n+1}\right)$ 2.$\sum_{2}^{\infty}\ln(\left(\frac{n+1}{n-1}\right)$

First I thought both of them are convergent as they are similar to $\sum \ln(n/n)$ and this series is just $0$+$0$..+$0$ and so by limit comparison test both of them converge.

But then when I treat above two as telescoping series I think they must be divergent.

Please clarify me.

  • $\begingroup$ Both sums are divergent. $\endgroup$ – user499203 Nov 24 '17 at 18:36
  • 1
    $\begingroup$ Regarding why your first argument fail: $$\ln\frac{n}{n+1} = - \ln\frac{n+1}{n} = -\ln\!\left(1+\frac{1}{n}\right)$$ and since $\lim_{x\to 0}\frac{\ln(1+x)}{x} =1$, by comparison the first series behaves like $-\sum_{n}\frac{1}{n}$ -- which diverges. $\endgroup$ – Clement C. Nov 24 '17 at 18:42
  • $\begingroup$ Why you apply limit to $\ln(1+x)$/$x$ $\endgroup$ – Believer Nov 24 '17 at 19:36
  • $\begingroup$ Will you please explain a bit more..I really want to know how you arrive at conclusion $\endgroup$ – Believer Nov 24 '17 at 19:39
  • $\begingroup$ Im not tracking your logic, Clement C. Not sure what your limit has to do with the problem. $\endgroup$ – CogitoErgoCogitoSum Nov 24 '17 at 19:42

$\bullet$ Consider the $n$-th partial sum of both series.

$\bullet$ $\displaystyle S_n=\sum_{k=1}^n[\ln(k)-\ln(k+1)]=-\ln(n+1)\to -\infty$ as $n\to \infty$.

$\bullet$ $\displaystyle S_n=\sum_{k=2}^n\ln\left(\frac{k+1}{k-1}\right)>\sum_{k=2}^{n}\ln \left(\frac{k}{k-1}\right)=\ln(n)\to \infty$ as $n\to \infty$.


1)$\log( \dfrac{n}{n+1}) = \log n - \log (n+1)$

$S_n:= \log(1) - \log(2) + \log(2) - \log(3)..+\log (n) -\log (n+1) =$

$ \log (1) - \log (n+1).$

2)$ \log (\dfrac{n+1}{n-1}) ={ \displaystyle \int_{n-1}^{n+1}\dfrac{1}{x}dx }$

$\gt 2 \dfrac{1}{n+1}.$

1) Find $\lim_{ n \rightarrow \infty} S_n.$

2) Any resemblance to the harmonic series?


Notice that the first sum can be written as $$\ln\left(\prod_{n=1}^\infty\frac{n}{n+1}\right)$$ Now, notice that all terms except $1$ and $n\to\infty$ cancel out, so we have $$\ln\frac1\infty=\ln0=-\infty$$ Therefore the sum is divergent. You can apply similar steps for the second sum too.

  • $\begingroup$ No need to go through products -- with sums, this is exactly what the OP referred to when they wrote "But then when I treat above two as telescoping series [...]." $\endgroup$ – Clement C. Nov 24 '17 at 18:43
  • $\begingroup$ @ClementC.. You've explained why his first argument fails. I just explained how these series can be solved using another approach. When it is written as a product, it is clear that it diverges. $\endgroup$ – user499203 Nov 24 '17 at 18:46
  • $\begingroup$ @GuyFsone. Arithmetic with infinity is allowed. Infinity is well-defined and we can use it in arithmetics. I really don't give a damn about votes, I'm not here for votes. $\endgroup$ – user499203 Nov 24 '17 at 18:58
  • $\begingroup$ @GuyFsone. "I had enough that day. I finally deleted my post." - Thank you for sharing your experience, I appreciate that, but not everyone agrees with the community. I'm not here for votes, I'm here to help others. If someone thinks that something is wrong in my post, that's really not my problem. And I'm not going to delete it if someone don't like it. $\endgroup$ – user499203 Nov 24 '17 at 19:05
  • $\begingroup$ @ThePirateBay..thanks for this method... $\endgroup$ – Believer Nov 24 '17 at 19:33

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