Convergence of series $\sum_{n=1}^\infty \ln\left(\frac{2n+7}{2n+1}\right)$?

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?

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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

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 ...?

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.

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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.

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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.

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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.

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@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 $x\leq 1$)

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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