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Show that $$\sum_{k\geq 2}\frac{(-1)^k(k-1)}{2k(k+1)n^k}<\frac{1}{12n(n+1)}$$ for all $n=1,2,\ldots$.

Its easy to see the series is convergent but i could not find an upper bound which is smaller than the right hand. Any idea? Thanks in advance.

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$$\begin{eqnarray*}\sum_{k\geq 2}\frac{(-1)^k}{n^k}\left(\frac{1}{k+1}-\frac{1}{2k}\right)&=&\int_{0}^{1}\sum_{k\geq 2}\frac{(-1)^k}{n^k}(2x^{2k+1}-x^{2k-1})\,dx\\&=&\int_{0}^{1}\frac{2x^5-x^3}{n(n+x^2)}\,dx\\&=&\frac{1}{2n}\int_{0}^{1}\frac{2z^2-z}{n+z}\,dz\\&=&\frac{1}{2n}\int_{-1/2}^{1/2}\frac{2z^2+z}{n+\frac{1}{2}+z}\,dz\\&=&\frac{1}{2n}\int_{0}^{1/2}\frac{16nz^2}{(2n+1)^2-4z^2}\,dz\\&\color{red}{<}&\frac{16n}{2n(2n)(2n+2)}\int_{0}^{1/2}z^2\,dz=\color{red}{\frac{1}{12n(n+1)}}.\end{eqnarray*}$$

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    Many thanks. This integral form is very nice. – 2017-02-09
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    @mudok: You're welcome. I guess this inequality is involved in a proof of Stirling's inequality. The integral representation also shows the bound is quite tight. – 2017-02-09