I have tried every test that we have learned so far in my class.

It's not geometric or telescoping, the Test for Divergence tells me nothing. It has negative terms, so I can't use the Integral Test, the Comparison Test, or the Limit Comparison test. The ratio test doesn't tell me anything.

Those are all the tests we've learned so far in my class. The other thing I've tried is separating it out using properties of logarithms:


However, those are two divergent series, and I can't find any theorem in my textbook that says that the sum of two divergent series is divergent.

How can I prove whether this series is convergent or divergent?

  • 1
    $\begingroup$ It is telescoping $\endgroup$
    – ziggurism
    Nov 17, 2016 at 4:39
  • $\begingroup$ Dang. I see it now. That took me way too long to figure out $\endgroup$
    – Connor M
    Nov 17, 2016 at 4:49
  • $\begingroup$ yes, it is telescoping (which is easier to use) but re the comparison test, all terms a negative, so if you look at $-\sum_{n=1}^\infty{\ln(\frac{n+1}{n+3})}=\sum_{n=1}^\infty(-\ln(\frac{n+1}{n+3}))=\sum_{n=1}^\infty{\ln(\frac{n+3}{n+1})}$ then all terms are positive. (Not that this observation helps, for the present series.) $\endgroup$
    – Mirko
    Nov 17, 2016 at 4:54
  • $\begingroup$ If you are interested only in convergence, you can also apply the limit comparison test. $\endgroup$
    – detnvvp
    Nov 17, 2016 at 5:01

3 Answers 3


Let $S_k := \sum_{n=1}^{k}\ln\Big(\frac{n+1}{n+3}\Big)$. By induction on $k$, $$\text{if }k\ge 3, \text{then }S_k =\ln\Big(\frac{6}{(k+2)(k+3)}\Big).$$

Hence, $\lim_{k \rightarrow \infty} S_k = -\infty $.


If you do not want to use the fact that this is telescoping, consider each one of the terms: $$-\ln\left(\frac{n+1}{n+3}\right)=-\ln\left(1-\frac{2}{n+3}\right).$$ Since $$\lim_{x\to 0}\frac{\ln(1+x)}{x}=1,$$ we obtain that $$\lim_{n\to\infty}\frac{-\ln\left(1-\frac{2}{n+3}\right)}{\frac{2}{n+3}}=1.$$ Therefore, the series $$\sum-\ln\left(\frac{n+1}{n+3}\right)$$ diverges, by comparison with the series $\sum\frac{2}{n+3}$. Hence, the original series diverges.


Given that $$ \ln(\frac{n+1}{n+3}) =$$ $$ =\ln(1+\frac{n+1}{n+3}-1) =$$ $$= \ln(1+\frac{n+1-n-3}{n+3}) =$$ $$= \ln(1+\frac{-2}{n+3}) \sim \frac{-2}{n+3} \sim -2 \frac{1}{n}$$, which diverges (it's the armonic series with $a$ equal to $1$). Hence, the given series diverges, by comparison with the series $$\sum_{n=1}^\infty{-2\frac{1}{n}}$$


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .