I have the series $ \small b = 1 + (1+ \frac 14) + (1 + \frac 14 + \frac 19) + \ldots $, where the partial sums are $$ s_2(n) = n + {n-1\over2^2} + {n-2\over 3^2} + \ldots + {1 \over n^2}$$ This series occurs in context with a generalization of the Eulerian numbers, which I've mentioned in this MO-thread. It looks, as if it is related to the simpler series $ \small a = 1 + \frac 12 + \frac 13 + \frac 14 + ... $whose partial sums are: $$ s_1(n) = 1 + {1\over2} + {1\over 3} + \ldots + {1 \over n}$$ where of course the limit as $ n \to \infty$ is infinite.
Question: can I relate the two (underlying) series (a,b) by some formal algebraic manipulation?
It might be, that the series with alternating signs can be related through the Dirichlet $\eta$- (or "alternating $\zeta$"), as shown in my linked answer in MO, but I don't see at the moment how I can support such a guess.
Note that $s_2(n)=(n+1)\sum\limits_{k=1}^n\frac1{k^2}-\sum\limits_{k=1}^n\frac1{k}=(n+1)\frac{\pi^2}6-(n+1)\sum\limits_{k=n+1}^\infty\frac1{k^2}-H_n$ with $H_n=\sum\limits_{k=1}^n\frac1{k}$ (and $H_n$ the $n$th harmonic number is also your $s_1(n)$).
Since $\sum\limits_{k=n+1}^\infty\frac1{k^2}=\frac1n+o\left(\frac1n\right)$ and $H_n=\log n+\gamma+o(1)$, one gets $$ \lim_{n\to\infty}\left(s_2(n)-n\frac{\pi^2}6+\log n\right)=\frac{\pi^2}6-1-\gamma. $$ Note that the constants involved are $\frac{\pi^2}6=\sum\limits_{k=1}^\infty\frac1{k^2}=\zeta(2)$ and $\gamma=\lim\limits_{n\to\infty}\left(\sum\limits_{k=1}^n\frac1k\right)-\log n$.