How to find the closed form of $\sum_{k=2}^{\infty}{\lambda(k)-1\over k}?$ Given that:
$$\sum_{k=2}^{\infty}{\lambda(k)-1\over k}\tag1$$
Where $\lambda(k)$ is Dirichlet Lambda Function
We are seeking to determine the closed form $(1)$ and came close to estimates it to $1-{\frac12}\left(\gamma+\ln{\pi}\right)$.
where $\gamma=0.5772156...$ is Euler-Mascheroni Constant
How can we evalauate the exact closed form for $(1)$?
 A: You have found it. Interchange the order of summation, and a few rewrites plus Wallis's product yield the result:
\begin{align}
\sum_{k = 2}^{\infty} \frac{1}{k}\bigl(\lambda(k) - 1\bigr)
&= \sum_{k = 2}^{\infty} \frac{1}{k} \sum_{n = 1}^{\infty} \frac{1}{(2n+1)^k} \\
&= \sum_{n = 1}^{\infty} \sum_{k = 2}^{\infty} \frac{1}{k(2n+1)^k} \\
&= \sum_{n = 1}^{\infty}\Biggl(-\frac{1}{2n+1} - \log \biggl(1 - \frac{1}{2n+1}\biggr)\Biggr) \\
&= \sum_{n = 1}^{\infty} \Biggl(\log \biggl(1 + \frac{1}{2n}\biggr) - \frac{1}{2n+1}\Biggr) \\
&= \sum_{n = 1}^{\infty} \Biggl(\frac{1}{2}\log \biggl(1 + \frac{1}{n}\biggr) - \frac{1}{2n} + \frac{1}{2n} - \frac{1}{2n+1} \\
&\qquad\quad - \frac{1}{2}\biggl(\log\biggl(1 + \frac{1}{n}\biggr) - 2\log \biggl(1 + \frac{1}{2n}\biggr)\biggr)\Biggr) \\
&= -\frac{1}{2}\gamma + 1 - \log 2 - \frac{1}{2} \log \prod_{n = 1}^{\infty} \frac{(2n+1)^2n}{(2n)^2(n+1)} \\
&= - \frac{1}{2}\gamma + 1 - \log 2 - \frac{1}{2} \log \Biggl(\prod_{k = 2}^{\infty}\biggl(1 - \frac{1}{k^2}\biggr)\prod_{m = 1}^{\infty}\biggl(1 - \frac{1}{(2m)^2}\biggr)^{-1}\Biggr) \\
&= - \frac{1}{2}\gamma + 1 - \log 2 - \frac{1}{2} \log \biggl( \frac{1}{2}\cdot \frac{\pi}{2}\biggr) \\
&= 1 - \frac{1}{2}\bigl(\gamma + \log \pi\bigr).
\end{align}
A: Another approach is to use the ordinary generating function of the Riemann zeta function, namely $$\sum_{n=2}^{\infty} \zeta(n) x^{n-1} = - \gamma - \psi (1-x), \quad |x| <1. $$
(This is just the Maclaurin series of the digamma function $\psi(1-x)$.)
Integrating both sides of the generating function, we get
$$\sum_{n=2}^{\infty} \frac{\zeta(n)}{n} \, x^{n} = -\gamma x + \log \Gamma(1-x) + C, $$ where the constant of integration must be zero since the value of the series is $0$ when $x=0$.
Therefore,
$$\begin{align} &\sum_{n=2}^{\infty} \frac{\lambda(n)-1}{n} \\ &= \sum_{n=2}^{\infty} \left(\frac{\zeta(n)}{n} - \frac{1}{n} \right) - \sum_{n=2}^{\infty} \frac{\zeta(n)}{2^{n}n} \tag{1}\\&= \lim_{x \to 1^{-}}\sum_{n=2}^{\infty} \left(\frac{\zeta(n)}{n} - \frac{1}{n} \right)x^{n} - \sum_{n=2}^{\infty} \frac{\zeta(n)}{2^{n}n} \tag{2}  \\ &=\lim_{x \to 1^{-}}\left({\color{red}{\sum_{n=2}^{\infty} \frac{\zeta(n)}{n}x^{n}}} - {\color{green}{\sum_{n=2}^{\infty}\frac{x^{n}}{n}}} \right) - \sum_{n=2}^{\infty} \frac{\zeta(n)}{2^{n}n} \\ &= \lim_{x \to 1^{-}} \big({\color{red}{\log \Gamma(1-x)- \gamma x}}  {\color{green}{ + \ln(1-x) + 1}} \big) + \frac{\gamma}{2} - \log \Gamma \left(1-\frac{1}{2} \right) \\ &=\lim_{x \to 1^{-}} \big(\log \Gamma(1-x) + \ln(1-x) \big)+1  -\frac{\gamma}{2}  - \frac{1}{2} \,  \log \pi  \\ &= \lim_{x \to 1^{-}} \log \big(\Gamma(1-x)\cdot (1-x) \big)+1 -\frac{\gamma}{2} - \frac{1}{2} \,  \log \pi \\  &= \lim_{x \to 1^{-}} \log \Gamma(2-x) +1  -\frac{\gamma}{2} - \frac{1}{2} \,  \log \pi  \\ &=0 +1 -\frac{\gamma}{2}  - \frac{1}{2} \,  \log \pi \\ &= 1 - \frac{1}{2} \left(\gamma + \log \pi \right) \end{align}$$

$(1)$ $\lambda(s) = (1-2^{-s}) \zeta(s)$
$(2)$ Abel's theorem
