We have the sequence of functions: $f_n(x) = \frac{1}{(1 + nx)(n + x)} \space $ on $I = ]0, + \infty[$.
We note: $f(x) = \sum_{n = 0}^{+ \infty} f_n (x)$.
- Examine the convergence and uniform convergence of $f_n$ on $ I$.
- Prove that the series $\sum f_n$ converges on $I$.
- Prove that the series $\sum f_n$ has normal convergence on the interval $[a, + \infty[$ with $ a > 0$ and not on $I$.
- Prove that $f$ is continuous on $I$.
- Prove that $f$ is integrable on $I$ and: $$ \int_{0}^{+ \infty} f(x)dx = 1 + 2 \sum_{n = 2}^{+ \infty} \frac{ln(n)}{n^2- 1}$$
- We have: $|f_n(x)| \leq \frac{1}{n^2 x}$
And: $ \sum |f_n(x)| \leq \sum \frac{1}{n^2 x} \leq \frac{1}{n^2} $ only if $ x \in [1, + \infty [$
since $ \sum \frac{1}{n^2}$ is convergent, so is $\sum ||f_n(x)||_{\infty}$ for $x \geq 1$. I am not sure about this. Is it correct?
From question 3, the series $\sum f_n$ converges normally on the interval $I = [a, + \infty[$, which means we have uniform convergence on the same interval. I can switch the sum and the integral.
$\begin{align} \int_{0}^{+ \infty} f(x)dx = & \int_{0}^{+ \infty} \sum_{n = 0}^{+ \infty} f_n(x)dx \\ & = \sum_{n = 0}^{+ \infty} \int_{0}^{+ \infty} f_n(x)dx \\ & = \int_{0}^{+ \infty} f_0(x)dx + \int_{0}^{+ \infty} f_1(x)dx + \sum_{n = 2}^{+ \infty} \int_{0}^{+ \infty} f_n(x)dx \end{align}$
I have problem with: $ \int_{0}^{+ \infty} f_0(x)dx = \int_{0}^{+ \infty} \frac{1}{x}dx$
I don't know how to compute this improper integral. If it is $0$, I will have the wanted result.
Although, if the series $\sum f_n$ has uniform convergence on the interval $[a, + \infty[$. Why do we take the integral from $0$? Why not $a$ ?
Thank you.