<s>[INST] Help for series calculation $\sum_{n\ge1} \frac{1}{4n^3-n}$ I want to find the following series.
$$\sum_{n\ge1} \frac{1}{4n^3-n}$$ 
So, fisrt of all, patial fraction : $$\frac{1}{4n^3-n}=\frac{1}{2n+1}+\frac{1}{2n-1}-\frac{1}{n}.$$
Next, I consider the geometric series $$\sum_{n\ge1} x^{2n-2}+x^{2n}-x^{n-1}=\frac{1}{1-x^2}+\frac{x^2}{1-x^2}-\frac{1}{1-x}\tag{1}$$ where ${-1\lt x\lt1}$.
Then, I integrate both side of (1) from $0$ to $1$:
$$\sum_{n\ge1}\frac{1}{2n-1}+\frac{1}{2n+1}-\frac{1}{n}=\int_0^1 \frac{x^2-x}{1-x^2}dx$$
Finally, I get the value that is equal to $ln(2)-1$  by improper integral but it's less than $0$. What did I do wrong? All help is appreciated.
 [/INST] You can do this way. It is easier. Let
$$ f(x)=\sum_{n=1}^\infty\frac{1}{4n^3-n}x^{2n+1}. $$
Then $f(1)=\sum_{n=1}^\infty\frac{1}{4n^3-n}$ and
$$ f'(x)=\sum_{n=1}^\infty\frac{1}{(2n-1)n}x^{2n}, f''(x)=2\sum_{n=1}^\infty\frac{1}{(2n-1)}x^{2n-1}, f'''(x)=2\sum_{n=1}^\infty x^{2n-2}=\frac2{1-x^2}$$
and hence
$$ f(1)=\int_0^1\int_0^s\int_0^t\frac2{1-x^2}dxdtds=2\ln2-1.$$
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