How can I solve this sequence with Riemann integration?

Question

I have this sequence defined by an improper integral $\displaystyle \lim_{n \to\infty} \int_0^\infty \cfrac{1}{1+x^2} \log\left(\frac{2nx+3}{nx+1}\right)\ dx$

it is easy to prove that this sequence converges with Lebesgue's integrations, but how can I prove the pointwise convergence with Riemann?

Answer 1

Use equivalence: $$\log\frac{2nx+3}{nx+1}\sim_{x\to\infty}\log\frac{2nx}{nx}=\log 2, \quad \text{hence }\;\frac1{1+x^2}\log\frac{2nx+3}{n+1}\sim_{x\to\infty} \frac{\log 2}{1+x^2},$$ which converges.

Answer 2

$$\log\frac{2nx+3}{nx+1} = \log\frac{2nx+2+1}{nx+1} = \log(2) + log(1+\frac{1}{2nx+2})$$

You can the use dominated convergence theorem since:

For all $x \in \mathbb{R}, $n in N:

$$\log(1+\frac{1}{2nx+2}) \le \frac{1}{2nx+2} \le \frac{1}{2}$$