# 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!!

• Why didn't you accept an answer? – Viktor Glombik Dec 11 '18 at 12:09

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

$$\sum_{n=1}^\infty a_{f(n)} = \lim_{k \rightarrow \infty} S_k = {\color{red}{\frac32}}\ln (2)$$

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