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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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    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)$2011-12-15

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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 $x\in[\,0,a\,]$.