# Evaluating the sum

Can anybody evaluate the following sum for me? $$\sum\limits_{n=2}^\infty(-1)^n\left(\frac{\psi(n)}{n}-\frac{\Lambda(n)}{2n}\right),$$ where $\psi(n)$ is the Chebyshev function and $\Lambda(n)$ is the Von Mangoldt function.

I tried using the fact that Chebyshev function is summatory Von Mangoldt function but it did not help. I am only interested in the numerical value.

I tried to find an exact closed form for this sum, but one part is left as a fairly rapidly converging series. The original sum is not entirely convergent either but may be normalized to a convergent value. $$S= \sum_{k=2}^{\infty}{(-1)^k}\left(\frac{\psi(k)}{k} - \frac{\Lambda(k)}{2k}\right) = \sum_{k=2}^{\infty}\frac{(-1)^k}{k}\left(\psi(k) - \frac{\Lambda(k)}{2}\right)$$ This inner term becomes equal to the Normalized Chebyshev Function $\psi_0(k)$. As this has the closed form. $$\psi_0(x)=x-\sum_\rho{\frac{x^\rho}{\rho}}-\ln(2\pi)-\frac{1}{2}\ln\left(1-x^{-2}\right) \\ \frac{\psi_0(x)}{x}=1-\sum_\rho{\frac{x^{\rho-1}}{\rho}}-\frac{\ln(2\pi)}{x}-\frac{1}{2x}\ln\left(1-x^{-2}\right)$$ Where $\rho$ is a non-trivial zero of the Riemann Zeta Function. This implies that the initial sum is: $$S= \sum_{k=2}^{\infty}{(-1)^k}\frac{\psi_0(k)}{k} \\ S= \sum_{k=2}^{\infty}{(-1)^k}\left(1-\sum_\rho{\frac{k^{\rho-1}}{\rho}}-\frac{\ln(2\pi)}{k}-\frac{1}{2k}\ln\left(1-k^{-2}\right)\right)\\S= \sum_{k=2}^{\infty}{(-1)^k}-\sum_{k=2}^{\infty}{(-1)^k}\sum_\rho{\frac{k^{\rho-1}}{\rho}}-\sum_{k=2}^{\infty}{(-1)^k}\frac{\ln(2\pi)}{k}-\sum_{k=2}^{\infty}{(-1)^k}\frac{1}{2k}\ln\left(1-k^{-2}\right)$$ The second sum may be evaluated by interchanging the order of summation: $$\sum_{k=2}^{\infty}{(-1)^k}\sum_\rho{\frac{k^{\rho-1}}{\rho}}=\sum_\rho\frac{1}{\rho}\sum_{k=2}^{\infty}{(-1)^k}{k^{\rho-1}}\\=-\sum_\rho{\frac{\eta(1-\rho)-1}{\rho}}=\sum_\rho{\frac{1}{\rho}}=\frac{\gamma}{2}+1-\frac{1}{2}\ln(4\pi)$$ The third is just a form of the natural logarithm of 2: $$\sum_{k=2}^{\infty}{(-1)^k}\frac{\ln(2\pi)}{k}=\ln(2\pi)\sum_{k=2}^{\infty}\frac{(-1)^k}{k}=\ln(2\pi)(\ln(2)-1)$$ The last sum does not seem to have a nice closed form but gives a decent approximation of: $$\sum_{k=2}^{\infty}{(-1)^k}\frac{1}{2k}\ln\left(1-k^{-2}\right)\approx-0.05773249...$$ The first sum gives either a value of 1 or 0 depending on an even or odd partial sum, i.e. Grandi's Series. This makes the initial sum divergent. Grandi's Series has the normalization of $\frac{1}{2}=\eta(0)$, giving three possible sums.

As a lower bound: $$S_1=0-\frac{\gamma}{2} -1 +\frac{\ln(4\pi)}{2}-\ln(2\pi)(\ln(2)-1)-\sum_{k=2}^{\infty}{(-1)^k}\frac{1}{2k}\ln\left(1-k^{-2}\right)\\=-1-\frac{\gamma}{2} -\frac{\ln(\pi)}{2}-\ln(2\pi)\ln(2)-\sum_{k=2}^{\infty}{(-1)^k}\frac{1}{2k}\ln\left(1-k^{-2}\right)\\ \approx -1-\frac{\gamma}{2} -\frac{\ln(\pi)}{2}-\ln(2\pi)\ln(2)+0.05773249...\approx -.5293209...$$ For an upper bound: $$S_2=S_1+1\approx .40679021...$$ As the normalized sum: $$S_3=S_1+\frac{1}{2}\approx-.02932097...$$

The normalized value is the most correct one, as the other two are just divergent messes, but sadly the last natural logarithm sum doesn't seem to have a nice closed form.

Just partial answer. Let $$f(k)=\sum\limits_{n=2}^k(-1)^n\left(\frac{\psi(n)}{n}-\frac{\Lambda(n)}{2n}\right).$$ The values of $f$ are

• $f(500) = 0.4723238355$
• $f(1000) = 0.4690203884$
• $f(1500) = 0.4737698493$
• $f(2000) = 0.4692923722$
• $f(2500) = 0.4703526128$
• $f(3000) = 0.4708612774$
• $f(3500) = 0.4683767275$
• $f(4000) = 0.4693891317$
• $f(4500) = 0.4701923562$
• $f(5000) = 0.4704750068$
• ...

You're asking for $\lim_{k \rightarrow \infty} f(k)$, but I even don't know how to prove that this series is convergent or not.