<s>[INST] 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 $(1)$?
 [/INST] 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}
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