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? 
 A: 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.
A: 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.
A: 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 $x\leq 1$)
A: 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.
A: 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.
