Uniform convergence of $ U_n(x) = \sum_{n=0}^{+ \infty} (-1)^n \ln ( 1 + \frac{x}{1+ nx} ) $. We consider the series of functions:
$$U_n(x) = \sum_{k=0}^{n} (-1)^k \ln \left( 1 + \frac{x}{1+ kx} \right)  ,~ x \geq0.$$

Prove that $U_n$ is convergent.

Study the uniform convergence of $U_n$.

Study the normal convergence.

We consider $U(x) = \sum_{n=0}^{+ \infty} (-1)^n \ln ( 1 + \frac{x}{1+ nx} )$. Prove that $U(x)$ is of class $C^1$.

Compute $U'(x)$.




I have problem with question 2 and 4.
For question 2. I do not know how to answer this question since I do not know the value of the sum $U_n (x)$ to compute : $\lim sup |U_n(x) - l| $. How to know if the series have uniform convergence?
For question 4. $U(x)$ is of class $C^1$ means it is differentiable and its derivative is continuous. If $ \sum U_n$ converges uniformaly, $U$ will have the same properties, but how can I can prove that $U$ is of $C^1$ without computing $U'(x)$ first? I am confused because in the last question I am asked to compute $U'(x)$. Is there a way to deduce that from uniform convergence?
This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
$$U_{2m-1}(x) = \sum_{k=0}^m \ln \left( 1+ \frac{x}{1+2kx} \right) - \ln \left( 1+ \frac{x}{1+(2k+1)x} \right)$$ which (after some calculations) turns out to be
$$\sum_{k=0}^m \ln \left( 1+ \frac{x^2}{(1+2kx)(1+2(k+1)x)} \right)$$

As $m \to + \infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $\Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.

Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).

As for differentiability of $U(x)$, you have to apply this limit .
This is a series with alternating sign terms, so a good idea would be to split it into even and odd terms. For even $n$, say $n=2m$, we have
$$U_{2m-1}(x) = \sum_{k=0}^m \ln \left( 1+ \frac{x}{1+2kx} \right) - \ln \left( 1+ \frac{x}{1+(2k+1)x} \right)$$ which (after some calculations) turns out to be
$$\sum_{k=0}^m \ln \left( 1+ \frac{x^2}{(1+2kx)(1+2(k+1)x)} \right)$$

As $m \to + \infty$ this series is normally convergent, thus $U_{2m-1}(x)$ is uniformly convergent on $\Bbb R$. Now you should study the series $U_{2m}$ and conclude similarly.

Thus $U_n$ is uniformly convergent, although it is not normally convergent (since it is not absolutely convergent).

As for differentiability of $U(x)$, you have to apply this limit .