I am studying real analysis with Rudin's Principles of Mathematical Analysis and a packet of exercises made by George Bergman that you can see here. I am trying to do Exercise 3.7:4, this is a generalization of Theorem 3.27 of Rudin's book, which states that if $(a_n)$ is a decreasing sequence of positive terms, then $\sum a_n$ converges if and only if $\sum 2^k a_{2^k}$ converges.
The exercise consists in proving the equivalence of some statements for a sequences of increasing positive integers $(n_k)$. The conditions are:
- If $(a_n)$ is any decreasing sequence of positive terms, the series $\sum a_n$ converges if and only if $\sum (n_{k+1}-n_{k}) a_{n_k}$ converges (the convergence of the series can be tested by "sampling" the terms $a_{n_k}$).
- If $(a_n)$ and $(b_n)$ are decreasing sequences of positive terms such that $b_{n_k}=a_{n_k}$ for all k, then $\sum a_n$ converges if and only if $\sum b_n$ does.
- The set of ratios $\frac{n_{k+1}}{n_k}$ is bounded.
- $(n_k)$ has a subsequence $(n_{m_k})$ such that the set of ratios $\frac{n_{m_{k+1}}-n_{m_k}}{n_{m_{k}}-n_{m_{k-1}}}$ is bounded.
I am stuck trying to prove the implication from the second to the third condition. It is suggested that I prove it in contrapositive form. This is my attempt:
Suppose that the set of ratios $\frac{n_{k+1}}{n_k}$ is unbounded. Then we want to find decreasing sequences $(a_n)$ and $(b_n)$ with positive terms such that $b_{n_k}=a_{n_k}$. We will set their common value to $c_k$, so let $(c_k)$ be a decreasing sequence with positive terms. Then we put $$a_1=a_2=\cdots=a_{n_1}=b_1=b_2=\cdots=b_{n_1-1}=c_1$$ $$a_n=c_{k+1} (n_k<n\le n_{k+1})$$ $$b_n=c_{k} (n_k\le n<n_{k+1})$$ This way, $(a_n)$ and $(b_n)$ are respectively the smallest and largest decreasing sequences starting with $c_1$ such that $a_{n_k}=b_{n_k}=c_k$. We want to choose $(c_k)$ that $\sum a_n$ converges while $\sum b_n$ diverges. We have
$$\sum_{n=1}^{n_m} a_n = (a_1+\cdots+a_{n_1})+(a_{n_1+1}+\cdots+a_{n_2})+\cdots+(a_{n_{m-1}+1}+\cdots+a_{n_m})$$ $$=n_1c_1 + (n_2-n_1)c_2 + \cdots + (n_m-n_{m-1})c_m=n_1c_1 + \sum_{k=1}^{m-1} (n_{k+1}-n_k)c_{k+1}$$
$$\sum_{n=1}^{n_m-1} b_n = (b_1+\cdots+b_{n_1-1})+(b_{n_1}+\cdots+b_{n_2-1})+\cdots+(b_{n_{m-1}}+\cdots+b_{n_m-1})$$ $$=(n_1-1)c_1 + (n_2-n_1)c_1 + \cdots + (n_m-n_{m-1})c_{m-1}=(n_1-1)c_1 + \sum_{k=1}^{m-1} (n_{k+1}-n_k)c_k.$$
Thus, we will be done if we can find a decreasing positive sequence $(c_k)$ such that $\sum(n_{k+1}-n_k)c_{k+1}$ converges but $\sum(n_{k+1}-n_k)c_k$ diverges.
I tried choosing $c_k = \frac{1}{n_k} p_k$ for some sequence $(p_k)$ such that $\frac{n_{k+1}}{n_k} p_k \ge p_{k+1}$, because this way $$\sum_{k=1}^m(n_{k+1}-n_k)c_{k+1}=\sum_{k=1}^m\frac{n_{k+1}-n_k}{n_{k+1}}p_{k+1}\le\sum_{k=1}^m p_{k+1}$$ $$\sum_{k=1}^m(n_{k+1}-n_k)c_k=\sum_{k=1}^m\frac{n_{k+1}-n_k}{n_k}p_k = \sum_{k=1}^m\frac{n_{k+1}}{n_k}p_k - \sum_{k=1}^m p_k$$.
Hence, with a sequence $(p_k)$ having $\frac{n_{k+1}}{n_k} p_k \ge p_{k+1}$ such that $\sum p_k$ converges but $\sum\frac{n_{k+1}}{n_k}p_k$ diverges, we would be done. But I can't think how to get such a sequence using only the fact that the ratios $\frac{n_{k+1}}{n_k}$ are unbounded, and I am not sure if my approach after defining $(a_n)$ and $(b_n)$ is right.
I am not looking for a solution, but any hint on how to continue would be greatly appreciated.
