Prove $\sum\limits_{n=1}^{\infty} a_{f(n)} = \dfrac{3}{2} \ln(2) $ Where $a_n = \dfrac{(-1)^{n+1}} { n}$ and  $f:\mathbb{N}\to \mathbb{N} $ is a bijection which for $k \in \mathbb{N}\setminus \{0\}$  is given by:  $$f(3k+1)= 4k+1$$ $$f(3k+2)= 4k+3$$ $$f(3k+3)= 2k+2.$$
Could some please help me get started on this proof? I'm having extra difficulty seeing where $f$ is constructed from (I've looked at various proofs of convergence of alternating harmonic series and am still confused). And then how to set up $a_{3n}$ based on this definition (again confused and didn't find that much step-by-step help on google). Thank you!!
 A: Euler's constant:
The sequence
$$\gamma_n := 1+\frac 12 + \frac 13 + \cdots + \frac 1n - \ln(n)$$
converges to a finite limit $\gamma$, where $\gamma$ is a constant:
$$\lim_{n \rightarrow \infty} \gamma_n = \gamma$$

The subsequence lemma:
The sequence $b_n$ converges to a limit $L$ if and only if all of the subsequences $b_{kn}$, $b_{kn+1}$, $b_{kn+2}$, $\dots$, $b_{kn+k-1}$ converge and have the same limit $L$.

Solution:
Define the sequence of partial sums
$$S_k = \sum_{n=1}^ka_{f(n)} = 1+\frac 13 - \frac 12 + \frac 15 + \frac 17 - \frac 14 + \cdots + a_{f(k)}$$
Let us first look at the subsequence $T_k := S_{3k}$.
\begin{align}
T_k = S_{3k} = & \bigg(1+\frac 13 - \frac 12\bigg) + \bigg(\frac 15 + \frac 17 - \frac 14\bigg) + \cdots + \bigg(\frac{1}{4k-3}+\frac{1}{4k-1}-\frac{1}{2k}\bigg) \\
= & \bigg(1+\frac 13 + \frac 15 + \cdots + \frac{1}{4k-1} \bigg)-\bigg(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac{1}{2k} \bigg) \\
= & \bigg[\bigg(1+\frac 12 + \frac 13 + \cdots + \frac{1}{4k-1}\bigg) - \bigg(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac{1}{4k-2}\bigg) \bigg] \\
& - \bigg(\frac 12 + \frac 14 + \frac 16 + \cdots + \frac{1}{2k} \bigg) \\
= & \bigg(1+\frac 12 + \frac 13 + \cdots + \frac{1}{4k-1}\bigg) - \frac 12 \bigg(1+\frac 12 + \frac 13 + \cdots + \frac{1}{2k-1}\bigg) \\
& -\frac 12 \bigg(1+\frac 12 + \frac 13 + \cdots + \frac{1}{k}\bigg) \\
= & [\gamma_{4k-1}+\ln(4k-1)]-\frac 12 [\gamma_{2k-1}+\ln(2k-1)] - \frac 12 [\gamma_{k}+\ln(k)] \\
= & \gamma_{4k-1}-\frac 12 \gamma_{2k-1} - \frac 12 \gamma_{k} + \frac 12 \ln\frac{(4k-1)^2}{(2k-1)k} \\
\rightarrow & \gamma - \frac 12 \gamma - \frac 12 \gamma + \frac 12 \ln (8) \\
= & {\color{red}{\frac32}}\ln (2)
\end{align}
Thus, $\lim_{k \rightarrow \infty}S_{3k} = {\color{red}{\frac32}}\ln (2)$. Of course, this is not enough to show that $\lim_{k\rightarrow \infty}S_k ={\color{red}{\frac32}} \ln (2)$ (take the Grandi series as a counter-example). We also need to show that $\lim_{k\rightarrow \infty}S_{3k+1} = \lim_{k\rightarrow \infty}S_{3k+2} = {\color{red}{\frac32}}\ln (2)$.
This is easily done with
$$\lim_{k\rightarrow \infty}S_{3k+1} = \lim_{k\rightarrow \infty}\bigg(S_{3k}+\frac{1}{4k+1} \bigg) = \lim_{k\rightarrow \infty}S_{3k}+\lim_{k\rightarrow \infty}\bigg(\frac{1}{4k+1} \bigg) = {\color{red}{\frac32}}\ln (2)+0 = {\color{red}{\frac32}}\ln (2)$$
and similarly,
$$\lim_{k\rightarrow \infty}S_{3k+2} = \lim_{k\rightarrow \infty}\bigg(S_{3k}+\frac{1}{4k+1}+\frac{1}{4k+3} \bigg) = {\color{red}{\frac32}}\ln (2)+0+0 = {\color{red}{\frac32}}\ln (2)$$
It follows that
$$\sum_{n=1}^\infty a_{f(n)} = \lim_{k \rightarrow \infty} S_k = {\color{red}{\frac32}}\ln (2)$$
A: Have you heard about so called Riemann series theorem? I think the proof contains a method of building an arbitrary sum with the permutation of conditionally convergent series and this idea could help you
