Is this way for solving convergence valid? $$\int_{1}^{\infty}\frac{\ln (x+1)-\ln(x)}{x}$$
solving:
$\int_{1}^{\infty}\frac{\ln ((x+1)/x)}{x}$
$\int_{1}^{\infty}\frac{\ln\sqrt[x]{(1+\frac{1}{x})^{x}}=e}{x}$
$\int_{1}^{\infty}\frac{\ln e^\frac{1}{x}}{x}$
$\int_{1}^{\infty}\frac{1}{x^{2}}$
from last line we can see that the intgral converges...
however the lecturer has written on the test that thr second line is wrong and that "$\int_{1}^{\infty}\frac{1}{x^{2}}\neq$" to the original intgral
I didn't get any points for this particular exercise.
why is the transition in second line is wrong(which is why in his opinion the whole answer is defect)? I've just written the term in another way so I can get 'e' I didn't change it in any way? Have I done something wrong?
The answer is that it is convergnet which is also what I got...
 A: No, your method is not valid because you have essentially claimed that for all $x \in [1, \infty)$, $$\frac{x+1}{x} = e^{1/x}.$$  Although when $x$ is "large," both the LHS and RHS tend to $1$, it is not justified when $x$ is near $1$.  The error occurs when you assert $$(1 + 1/x)^x = e.$$  This is only true as a limit.
If, however, you had found a function $g(x) \ge \frac{\log (x+1) - \log x}{x}$ on $[1, \infty)$ such that $$\int_{x=1}^\infty g(x) \, dx < \infty$$ then this would assure convergence of the original integral.

Although not in the scope of your question--which merely asks for whether the integral converges--the integral can be evaluated via series expansion about $\infty$:  $$\frac{\log (x+1) - \log x}{x} = x^{-1} \cdot \log (1 + x^{-1}) = \sum_{k=2}^\infty \frac{(-1)^k}{(k-1) x^k}.$$  Therefore $$\int_{x=1}^\infty \frac{\log (x+1) - \log x}{x} \, dx = \sum_{k=2}^\infty \int_{x=1}^\infty  \frac{(-1)^k}{(k-1) x^k} \, dx = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k^2}.$$  This sum has value $\pi^2/12$ via an elementary modification of the solution to the Basel problem.
A: In fact, your assertion is wrong.
$$
\int_1^\infty\frac{\log(x+1)-\log(x)}{x}\;dx = \frac{\pi^2}{12}
\tag{1}$$
but
$$
\int_1^\infty \frac{1}{x^2}\;dx = 1
\tag{2}$$
As noted, a correct way to solve this would be to show
$$
\frac{\log(x+1)-\log(x)}{x} \le \frac{1}{x^2}
$$
and then use the fact that $(2)$ is finite.
