Divergence of a simple improper integral $\int_{0}^{\infty} \ln \left(\frac{x+2}{x+1} \right)dx$ I have a reduced a problem I have been working on to showing that $$\int_{0}^{\infty} \ln \left(\frac{x+2}{x+1} \right)dx$$
diverges, but I'm not sure how to show this. Would anyone be able to help me out?
 A: By the mean value theorem $\ln\left(\frac{x+2}{x+1}\right)=\ln(x+2)-\ln(x+1) = \frac{1}{c(x)}$, where $x+1< c(x)<x+2$. Thus $\ln\left(\frac{x+2}{x+1}\right)\geq \frac{1}{x+2}$. 
Since the integral $\int_{1}^\infty \frac{1}{x+2} dx$ diverges, we get that $\int_{1}^\infty \ln\left(\frac{x+2}{x+1}\right)dx$ diverges, and hence $\int_{0}^\infty \ln\left(\frac{x+2}{x+1}\right)dx$ diverges.
A: First note $\ln\left(\dfrac{x+2}{x+1} \right)$ is a decreasing function. Hence, $$\int_{n-1}^{n} \ln\left(\dfrac{x+2}{x+1} \right) dx > \ln \left(\dfrac{n+2}{n+1} \right)$$
Hence,
\begin{align}
\int_0^{N} \ln\left(\dfrac{x+2}{x+1} \right) dx & = \sum_{n=1}^{N} \int_{n-1}^{n} \ln\left(\dfrac{x+2}{x+1} \right) dx\\
& > \sum_{n=1}^{N} \ln \left(\dfrac{n+2}{n+1} \right)\\
& = \sum_{n=1}^{N} \left(\ln(n+2) - \ln(n+1) \right)\\
& = \ln(N+2) - \ln(2)
\end{align}
Now you should be able to finish it off by letting $N \to \infty$.
A: Another approach may arise as following:
Let $g(x)=\frac{1}{x+1}$. Then it is non-negative and $$\lim_{x\to +\infty}\frac{\ln\frac{x+2}{x+1}}{\frac{1}{x+1}}=1$$ which is not zero or $\infty$; so the Quotient test says, $\int_0^{\infty}\ln\frac{x+2}{x+1} dx $ and $\int_0^{\infty}\frac{1}{x+1} dx $ are the same in being divergence or in being convergence.  The second integral is clearely divergent one.
A: A pedestrian approach:
\begin{eqnarray*}
\int_{0}^{n} \ln \left(\frac{x+2}{x+1} \right)dx
&=& \int_{0}^{n} \ln \left(x+2\right)dx
    -\int_{0}^{n} \ln \left(x+1\right)dx \\
&=& \ldots \\
&=& \log\frac{(n+2)^{n+2}}{4(n+1)^{n+1}} \\
&=& \log n + O(1) 
    \hspace{5ex} (\textrm{expand in large }n)
\end{eqnarray*}
Above we use the standard integral
$$\int\log(t)dt = t\log t - t.$$
