my motivation is to find a "closed form" (may contain integrals) of the general infinite sum $$\sum_{n=0}^\infty\frac{1}{(2n+2k+1)(4n+2k+3)}$$, where k is an arbitrary natural number. The sum can be rewritten as $$\frac{1}{1-2k}\sum_{n=0}^\infty\frac{1-2k}{(2n+2k+1)(4n+2k+3)}$$which is in turn equals$$\frac{1}{1-2k}\sum_{n=0}^\infty\frac{(4n+2k+3)-2(2n+2k+1)}{(2n+2k+1)(4n+2k+3)}=$$
$$\frac{1}{1-2k}\sum_{n=0}^\infty(\frac{1}{(2n+2k+1)}-\frac{2}{(4n+2k+3)})$$The individual terms of the sum can be rewritten using integrals$$\frac{1}{1-2k}\sum_{n=0}^\infty\int_0^1(x^{2n+2k}-x^{4n+2k+2})dx$$Now one could swap the summation and integration signs to obtain$$\frac{1}{1-2k}\int_0^1\sum_{n=0}^\infty(x^{2n+2k}-2x^{4n+2k+2})dx$$
As a test I have put both the original and this expression into Desmos (with 10000 terms and k=1) and both sums appear to be equal to approx. 0.132, so it seems that in this case such a swap is allowed. The thing is, one could use the Taylor series $$\sum_{n=0}^\infty{x^{2n}}=\frac{1}{1-x^{2}}$$ and $$\sum_{n=0}^\infty{x^{4n}}=\frac{1}{1-x^{4}}$$to simplify it even further to $$\frac{1}{1-2k}\int_0^1(\frac{x^{2k}}{1-x^2}-\frac{2x^{2k+2}}{1-x^4})dx$$, but for some reason, this integral is equal to -0,215 (according to Desmos), which obviously doesn't make sense, since the original sum must be positive. However, I don't know where I made the mistake in the conversion from Taylor series to $$\frac{1}{1-x^{n}}$$