Prove: $\sum \limits_{n=1}^{\infty}((n+\frac{x}{n})\ln(1+\frac{x}{n})-x)$ for $x \in [0,a]$ uniformly converges

I'd like your help with proving that following sum uniformly converges:

$$\sum_{n=1}^{\infty}\left(\left(n+\frac{x}{n}\right)\ln\left(1+\frac{x}{n}\right)-x\right)$$ for $x \in [0,a]$.

I tried to use the theorem saying that if there's one point (in our case $x=0$), which the series pointwise converges, and the sum of $U'_n$ converges also, the original sum uniformly converges, but it didn't work for me here.

Any hints?

Thanks a lot!

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What is $U'_n$? – Dirk Dec 15 '11 at 20:12
It's the derivative of $U_n$ when $U_n = \sum_{n=1}^{\infty}\left(\left(n+\frac{x}{n}\right)\ln\left(1+\frac{x}{n}\right)‌​-x\right)$ – Jozef Dec 15 '11 at 20:28

The series does not converge if $x\ne0$. Using the Taylor expansion $$\ln(1+t)=t-\frac{t^2}{2}+\cdots$$ we see that $$\Bigl(n+\frac{x}{n}\Bigr)\ln\Bigl(1+\frac{x}{n}\Bigr)-x=-\frac{x^2}{2\,n}+O\Bigl(\frac{1}{n^2}\Bigr)$$ where the $O$ term is uniform in $x\in[\,0,a\,]$.