Help for series calculation $\sum_{n\ge1} \frac{1}{4n^3-n}$

Following a comment of mine above:

• without checking too carefully, the issue in your step most likely resides in the "I integrate both sides of (1) from 0 to 1." You then integrate the LHS termwise -- why can you do that? The interval of convergence of your power series is $1$, so it is not obvious you can swap $\sum_n$ and $\int_0^1$. And, as it turns out, worse than "not obvious" it appears to be wrong.

• Now, another approach to compute the sum: starting with your $(1)$, we write, for $N\geq 1$, $$\begin{align} \sum_{n=1}^N a_n &= \sum_{n=1}^N \frac{1}{2n+1}+\sum_{n=1}^N \frac{1}{2n-1}-\sum_{n=1}^N \frac{1}{n}\\ &= \sum_{n=1}^N \left(\frac{1}{2n+1}+\frac{1}{2n}\right)+\sum_{n=1}^N \left(\frac{1}{2n-1}+\frac{1}{2n}\right)-2\sum_{n=1}^N \frac{1}{n} \\ &= \sum_{n=2}^{2N+1} \frac{1}{n}+\sum_{n=1}^{2N} \frac{1}{n}-2\sum_{n=1}^N \frac{1}{n} \\ &= H_{2N+1}-1+H_{2N}-2H_N \\ &= \ln(2N+1)+\ln(2N)-2\ln N - 1 +o(1)\\ &= 2\ln 2 - 1 + \ln(N+\frac{1}{2})+ \ln(N+\frac{1}{2})-2\ln N +o(1)\\ &= 2\ln 2 - 1 + \ln\frac{N+\frac{1}{2}}{N}+ \ln\frac{N-\frac{1}{2}}{N} +o(1)\\ &\xrightarrow[N\to\infty]{} \boxed{2\ln 2 -1} \end{align}$$ where $H_N = \sum_{n=1}^N \frac{1}{n}$ denotes the $N$-th Harmonic number, which satisfies $$ H_N = \ln N + \gamma + o(1) $$ when $N\to\infty$ ($\gamma$ being Euler's constant). We also used that by continuity of the logarithm, $\ln\frac{N\pm\frac{1}{2}}{N} \xrightarrow[N\to\infty]{} \ln 1=0$.

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 $$ f(1)=\int_0^1\int_0^s\int_0^t\frac2{1-x^2}dxdtds=2\ln2-1.$$