Prove that $f(x)=\sum_{n=1}^{\infty} (\frac{x}{n}-log(1+\frac{x}{n}))$ is continuous and can be differentiated ad infinitum We have $f:(0,\infty) \rightarrow \mathbb{R}$ defined by infinite series $f(x)=\sum_{n=1}^{\infty}  (\frac{x}{n}-log(1+\frac{x}{n}))$
Prove that $f$ is continuous and can be differentiated infinitely.
I have problem right in the beginning. I do not know how to prove that above series in uniformly convergent. I was trying to use series expansion of natural logarithm, but I get this $f(x)=\sum_{n=1}^{\infty}\sum_{k=2}^{\infty}  \frac{(-1)^kx^k}{kn^k}$ and I don't know how to proceed.
 A: By "$\log$" you mean the "natural logarithm".
From Wiki of "Natural logarithm", we have the following inequality for all $x\in(0,\infty)$,
$$\frac{x}{x+1}\le\log(1+x)\le x.$$
Then we see that each term $u_n(x):=\frac{x}{n}-\log(1+\frac{x}{n})$ satisfies
$$0\le u_n(x)\le\frac{x}{n}-\frac{\frac{x}{n}}{\frac{x}{n}+1}=\frac{x^2}{nx+n^2}.$$
Now for $\forall M>0$ and $\forall x\in(0,M]$, we have
$$0\le u_n(x)\le\frac{M^2}{n^2},$$
and then
$$0\le f(x)=\sum_{n=1}^\infty u_n(x)\le M^2\bigg(\sum_{n=1}^\infty\frac{1}{n^2}\bigg).$$
Hence by Weierstrass criterion,  $f$ converges uniformly on $(0,M]$ for all $M>0$, and thereby converges (but may not uniformly) on $(0,\infty)$ by the arbitrariness of $M$.
Therefore, the continuity of $f$ follows from this property of each term $u_n$ and the uniform convergence of $f$ on $(0,M]$ for all $M>0$.
Similarly, for $\forall M>0$ and $\forall x\in(0,M]$,
$$0\le u_n'(x)=\frac{1}{n}-\frac{1}{n+x}\le\frac{M}{n^2},$$
$$0\le|u_n^{(k)}(x)|=\bigg|\frac{(-1)^k}{(n+x)^k}\bigg|\le\frac{1}{n^k},$$
thus the infinite differentiability of $f$ follows from the continuity and infinite differentiability of each term $u_n$ and the uniform convergence of all-order derivatives $u_n^{(k)}$ on $(0,M]$ for all $M>0$.
