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.
 A: 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.
A: 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.
