Prove that $\sum \alpha_n$ converges $\iff$ $\sum (n_{k+1}-n_k)\alpha_{n_k}$ converges.

Question

Let $(\alpha_n)_{n\in \mathbb{N}}$ be a monotonically-decreasing sequence of positive reals.
Suppose that $(n_k)_{k\in\mathbb{N}}$ is a subsequence of $(n)_{n\in \mathbb{N}}$ with the property,
$\exists M\in \mathbb{R} \space \space \space\forall k \in \mathbb{N } : n_{k+1} - n_{k} \leq M(n_k- n_{k-1})$

Prove that $\sum \alpha_n$ converges $\iff$ $\sum (n_{k+1}-n_k)\alpha_{n_k}$ converges.

If we assume $\sum \alpha_n$ converges, we can deduce $\sum \alpha_{n_k}$ converges.
I'm still not sure what would be the next step. Moreover, I observed that for each $k$, $\sum_{j=1}^{k} n_{j+1} - n_{j_j} = n_{k+1}-n_1$, which perhaps suggest to use Abel's summation in a way which I struggle to figure.

Would like to have some insights, thanks

You can refer to the proof of Theorem $1$ in the link provided in Prove that if $a_1 + a_2 + \ldots$ converges then $a_1+2a_2+4a_4+8 a_8+\ldots$ converges and $\lim na_n=0$. This is known as the Schlömilch condensation test, which is a generalisation of Cauchy's condensation test. — Bruno B, Aug 10 '23 at 13:44

score 2 · Accepted Answer · answered Aug 10 '23 at 13:49

Since $(a_i)$ is non-increasing, one has $$ (n_{k+1}-n_k)a_{n_k}\geqslant \sum_{i=n_k+1}^{n_{k+1}}a_i. $$ Moreover, using the assumption on $(n_k)$, $$ a_{n_k}(n_{k+1}-n_k)\leqslant Ma_{n_k}(n_k-n_{k-1})\leqslant M\sum_{i=n_{k-1}+1}^{n_k} a_i $$ Denoting $S_N:=\sum_{i=n_0+1}^Na_i$, one has, in view of the previous inequalities, that $$ S_{n_{K+1}}\leqslant\sum_{k=1}^Ka_{n_k}(n_{k+1}-n_k)\leqslant MS_{n_K}. $$

Prove that $\sum \alpha_n$ converges $\iff$ $\sum (n_{k+1}-n_k)\alpha_{n_k}$ converges.

1 Answers1