Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have the series $$\sum\limits_{n=1}^\infty \ln\left(\frac{2n+7}{2n+1}\right)$$

I'm trying to find if the sequence converges and if so, find its sum.

I have done the ratio and root test but It seems it is inconclusive.

How can I find if the series converges?

share|improve this question
Try finding a simple equivalent of the summand. –  Joel Cohen Jan 27 '13 at 19:24
Look for an equivalent of the general term. If you don't know this technique, use comparison and the inequality $\ln (1+u)\geq u-u^2/2$ for all $u\geq 0$. –  1015 Jan 27 '13 at 19:39

5 Answers 5

up vote 7 down vote accepted

You can use the comparison test:

$$\sum\limits_{n=1}^\infty \ln\left(\frac{2n+7}{2n+1}\right) \quad = \quad\sum_{n=1}^\infty \ln\left(1+\frac{6}{2n+1}\right)\quad \geq \quad \sum_{n=1}^\infty \frac{6}{2n+1}-\frac{6^2}{2(2n+1)^2}$$
As $\,n \to \infty,\,$ the right-hand sum $\to \infty\,$ (so the right-sum diverges). And so, by the comparison test, any sum greater than a divergent sum must ...?

Added, per comments below:

Step $(1) \to (2)$: polynomial division. Note that $\;\dfrac{2n+7}{2n+1} = \dfrac{(2n+1)+6}{2n + 1}.$

Step $(2) \to (3)$: The inequality uses the fact that the Taylor series of $\;\ln(1 + x)\;$ with $\;x = \dfrac{6}{2n+1}$, is given by $\;x - \dfrac{x^2}{2} + \dfrac{x^3}{3} ...$, so the inequality follows from the omission of all but the first two terms of this series.

share|improve this answer
Yes. it does works. –  Babak S. Jan 27 '13 at 19:28
Sorry but did not understand How you went from step one to step two. –  Favolas Jan 27 '13 at 19:54
@amWhy Sorry got to get away from the computer. Thanks for your explanation. I understand the step and understand why the sum diverges. I don't understand the Taylor series part because I didn't get to that part yet. Probably it will be clearer some time soon. Many many thanks –  Favolas Jan 27 '13 at 20:55
Your very welcome! –  amWhy Jan 27 '13 at 21:07

No need for convergence tests! Note that if $f(n)=\ln\left(\dfrac{2n+7}{2n+1}\right)$ then:

$$f(n)+f(n+3)=\ln (2n+13)-\ln (2n+1)$$

So most terms cancel out. In other words, the partial sum of your series is:

$$\begin{align*}\sum_1^m f(n)&=-\ln 3-\ln 5-\ln 7+\ln(2m+3)+\ln(2m+5)+\ln(2m+7)\\ &= \ln(2m+3)(2m+5)(2m+7)-\ln 105\end{align*}$$

Which obviously does not converge.

share|improve this answer
Wow, that is a much more elegant way to prove divergence than my post! Very cool. –  Peder Jan 27 '13 at 19:47
The problem with this method is that it depends on the summand having a very special form. The methods using $\ln(1+x) \approx x + O(x^2)$ as $x \to 0$ work for any summand of the form $(an+b)/(an+c)$. –  marty cohen Jan 28 '13 at 3:27
@marty So? Every problem is different - if something has a special form, why not make use of it? Seems rather dull to stick to certain methods simply because they are more practical in general... –  L. F. Jan 28 '13 at 3:32

We will use the equivalent $$ \ln (1+u)\sim u $$ as $u$ approaches $0$.

The general term satisfies: $$ \ln\left(\frac{2n+7}{2n+1} \right)=\ln\left( 1+ \frac{6}{2n+1}\right)\sim \frac{6}{2n+1}\sim \frac{3}{n} $$ as $n$ approaches $+\infty$.

Now the series with general term $3/n$ diverges (cf Riemann series).

So the original series diverges.

share|improve this answer

Hint: Note that $$\lim_{n\to+\infty}\frac{\ln\left(\frac{2n+7}{2n+1}\right)}{n^{-1}}\neq0$$ so since the power of $n$ in the denominator is $-1$, so the series diverges.

share|improve this answer
@amWhy: You know, I have used this method for such these problems here. –  Babak S. Feb 2 '13 at 6:28

The sequence does not converge, because

$$ \sum_{n=1}^N \log\left(\frac{2n+7}{2n+1}\right) = \sum_{n=1}^N \log\left(1+\frac{6}{2n+1}\right)\geq \sum_{n=1}^N \frac{6}{2n+1}-\frac{36}{2(2n+1)^2} $$ and the sum on the right goes to infinity as $N\rightarrow\infty$. (The inequality follows by the Taylor series $\log(1+x)=x-x^2/2+x^3/3$ and the fact that this is an alternating series with decaying terms when $x\leq 1$)

share|improve this answer
I believe there should not be a $1$ in your right hand sum. –  1015 Jan 27 '13 at 19:29
I edited the post. Thanks for catching the mistake –  Peder Jan 27 '13 at 19:46
The first equality seems to be wrong...? –  DonAntonio Jan 27 '13 at 20:20
Check it now. I guess my ones got all bunched up in the wrong places... –  Peder Jan 27 '13 at 20:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.