Existence of a sequence related to the convergence of a series Trying to prove an exercise, I arrived at the following question:
Let $\{a_j\}\subseteq\mathbb{R}^+$ be a monotone increasing sequence with limit $+\infty$. Suppose that there is a $D>0$ such that $a_j\leq D j$ for all $j$. Do there exist a monotone increasing sequence $\{b_j\}\subseteq\mathbb{R}^+$ with limit $+\infty$ and a number $C>0$ such that $b_j\leq Ca_j$ and $$\sum_{j=1}^{\infty}(a_j^{-1}-a_{j+1}^{-1})b_j<\infty\,?$$
Any hint would be appreciable.
 A: The answer is yes for a very large class of sequences $\{a_j\}$. Let $a_j=f(j)$ where $f\colon[1,\infty)\to\mathbb{R}^+$ is a differentiable function that satisfies the following conditions:


*

*$\int_1^\infty f(x)\,dx=\infty$

*$0<f'(x)\le D\,x$, $D\ge f(1)$

*$f(x+1)-f(x)\le C\,x$

*$\dfrac{f'}{f^{1+p}}$ is decreasing for some $p\in(0,1)$


Then $\displaystyle\sum_{n=1}^\infty\Bigl(\frac{1}{a_j}-\frac{1}{a_{j+1}}\Bigr)\,(a_j)^{1-p}<\infty$.
From 1. and 2. we see that $a_j$ is increasing and converges to $\infty$ as $j\to\infty$, and that $a_j\le D\,j$. Moreover
$$
\frac{1}{a_j}-\frac{1}{a_{j+1}}=\frac{f(j+1)-f(j)}{f(j)\,f(j+1)}\le C\,\frac{f'(j)}{f(f)^2}
$$
Then
$$
\sum_{n=1}^\infty\Bigl(\frac{1}{a_j}-\frac{1}{a_{j+1}}\Bigr)\,(a_j)^{1-p}\le C\sum_{n=1}^\infty\frac{f'(j)}{f(j)^{1+p}}.
$$
The last series is convergent by the integral test, since $f'/f^{1+p}$ is nonnegative, decreasing and
$$
\int_1^\infty\frac{f'(x)}{f(x)^{1+p}}\,dx=-\frac1p\int_1^\infty\Bigl(\frac{1}{f^p}\Bigr)'\,dx=\frac{1}{p\,(a_1)^p}<\infty.
$$
Finally, since $a_j$ converges to $\infty$ and $0<p<1$, $a_j^p$ also converges to $\infty$ and $(a_j)^p/a_j=(a_j)^{-(1-p)}$ converges to $0$, and in particular is bounded.
Examples of functions $f$ are $x^p$, $0<p<1$, $\log x$, $\log\log x$,... 
A: The key idea is to use the following result about series' comparisons:
LEMMA: Let $\{\alpha_n\}_{n=1}^{\infty}\subseteq\mathbb{R}^+$ be a sequence with $\sum_{n=1}^{\infty}\alpha_n<\infty$. Then there exists another sequence $\{\beta_n\}_{n=1}^{\infty}\subseteq\mathbb{R}^+$ with $\sum_{n=1}^{\infty}\beta_n<\infty$ and $$\lim_n\frac{\beta_n}{\alpha_n}=\infty$$ in a monotone way.
Proof: Consider $\{A_k\}_{k=1}^{\infty}$ defined as $A_k=\sum_{n=k}^{\infty}\alpha_n$. Since $\lim_kA_k=0$, take a subsequence $\{A_{k_{l}}\}_{l=1}^{\infty}$ such that $A_{k_{l}}<2^{-l}$ for all $l\in\mathbb{N}$. Define $\beta_n=l\alpha_n$ if $n\in [k_l,k_{l+1}[$. 
Coming back to the question, the case in which $\{a_n\}_{n=1}^{\infty}$ is strictly increasing is easy. Consider $\alpha_n=a_n^{-1}-a_{n+1}^{-1}>0$. Since $\lim_na_n=\infty$, we have $\sum_{n=1}^{\infty}\alpha_n<\infty$. Take the corresponding $\{\beta_n\}_{n=1}^{\infty}$ from the lemma. Let $\hat{b}_n=\beta_n/\alpha_n$. Then $\{\hat{b}_n\}$ is increasing, tends to $\infty$ and $$\sum_{n=1}^{\infty}(a_n^{-1}-a_{n+1}^{-1})\hat{b}_n=\sum_{n=1}^{\infty}\beta_n<\infty.$$ Now define $b_n=\min\{\hat{b}_n,a_n\}$. Then $\lim_nb_n=\infty$, $\{b_n\}$ is increasing, $$\sum_{n=1}^{\infty}(a_n^{-1}-a_{n+1}^{-1})b_n\leq \sum_{n=1}^{\infty}(a_n^{-1}-a_{n+1}^{-1})\hat{b}_n<\infty$$ and $b_n\leq Ca_n$, with $C=1$.
If $\{a_n\}$ is increasing but not strictly, define $\alpha_n=a_n^{-1}-a_{n+1}^{-1}$ if $a_n<a_{n+1}$, $\alpha_n=2^{-n}$ if $a_n=a_{n+1}$. Then $\alpha_n>0$ and $$\sum_{n=1}^{\infty}\alpha_n\leq \sum_{n=1}^{\infty}(a_n^{-1}-a_{n+1}^{-1})+\sum_{n=1}^{\infty}2^{-n}<\infty.$$ Then as before one finds $\{\hat{b}_n\}$ such that it is increasing, it tends to $\infty$ and $\sum_{n=1}^{\infty}\alpha_n\hat{b}_n<\infty$. Then $\sum_{n=1}^{\infty}(a_n^{-1}-a_{n+1}^{-1})\hat{b}_n\leq \sum_{n=1}^{\infty}\alpha_n\hat{b}_n<\infty$. Now consider $b_n=\min\{\hat{b}_n,a_n\}$ and reason as before.
Notice that the hypothesis $a_n\leq Dn$ has not been used.
