I am interested in finding closed form solutions for the positive odd integers of the Riemann zeta function, of which only 1 is known. Please forgive me if this is already proven or well-known, but in running some calculations it appears that the resulting sum when the positive odd integer values of the Riemann zeta function are subtracted from the positive even integer values thereof, the result is ½; that is to say: $$\sum_{n=1}^{\infty} \biggl(\zeta(2n)-\zeta(2n-1)\biggr) = {1\over 2}$$ such that: $$=\biggl(\zeta(2)-\zeta(1)\biggr)+\biggl(\zeta(4)-\zeta(3)\biggr)+\biggl(\zeta(6)-\zeta(5)\biggr)+...+\biggl(\zeta(2n)-\zeta(2n-1)\biggr)$$ The positive even integer values of $\zeta(n)$ are well known, but no closed form expression is currently known for the positive odd integers of $n>1$, so I've included the decimal equivalents here: $$=\biggl(\frac{\pi^2}{6}-1\biggr)+\biggl(\frac{\pi^4}{90}-1.20205690315959428539\biggr)+\biggl(\frac{\pi^6}{945}-1.03692775514336992633\biggr)+\biggl(\frac{\pi^8}{9450}-1.00834927738192282683\biggr)$$ Performing this calculation, as $n$ increases, the sum approaches $\infty$.
If this is currently unproven and is true, it perhaps may lead to further revelations regarding the Riemann zeta function and/or finding closed form expressions for the positive odd values of $n$.
A possible corrected interpretation (to get rid of $\color{red}{\zeta(1)}=\infty$) is to consider the sum of $$\zeta(2n)-\zeta(2n+1)=\sum_{k=\color{red}{2}}^\infty(k^{-2n}-k^{-2n-1})=\sum_{k=2}^\infty(k-1)k^{-2n-1}$$ over $n>0$. After interchange of the summations, we get a telescoping series: $$\sum_{n=1}^\infty\big(\zeta(2n)-\zeta(2n+1)\big)=\sum_{k=2}^\infty\frac{k-1}{k^3}\left(1-\frac{1}{k^2}\right)^{-1}=\sum_{k=2}^\infty\frac{1}{k(k+1)}=\frac12.$$