Show that $\sum_{n=1}^{\infty}a_{n}$ converges iff $\sum_{k=1}^{\infty}\left(n_{k+1}-n_{k}\right)a_{n_{k}}$ converges

Question

Question

Let $\left(a_{n}\right)_{n=1}^{\infty}$ be a positive decreasing sequence in $\mathbb{R}$.

Let $\left(n_{k}\right)_{k=1}^{\infty}$ be a strictly increasing sequence in $\mathbb{N}$ so there exists $M\in\mathbb{R}$ such that: $$ n_{k+1}-n_{K}\leq M\left(n_{k}-n_{k-1}\right) $$ for every $k\in\mathbb{N}$.

Show that the series $\sum_{n=1}^{\infty}a_{n}$ converges iff the series $\sum_{k=1}^{\infty}\left(n_{k+1}-n_{k}\right)a_{n_{k}}$ converges.

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Edit 1

The first comment indicated to me that this is a generalization of Cauchy condensation test. I am currently checking its proof to look for similarities to this one but no success yet.

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Edit 2

My latest attempt goes like this (similar to the proof of the condensation test):

$$ \sum_{k=1}^{n}\left(n_{k+1}-n_{k}\right)a_{n_{k}}=\left(n_{2}-n_{1}\right)a_{n_{1}}+\left(n_{3}-n_{2}\right)a_{n_{2}}+\left(n_{4}-n_{3}\right)a_{n_{3}}+\left(n_{5}-n_{4}\right)a_{n_{4}}+\ldots $$ $$ \leq\left(n_{2}-n_{1}\right)a_{n_{1}}+M\left(n_{2}-n_{1}\right)a_{n_{2}}+M\left(n_{2}-n_{1}\right)a_{n_{3}}+M\left(n_{2}-n_{1}\right)a_{n_{4}}+\ldots $$ $$ \underbrace{=}_{\begin{array}{c} \text{Parentheses insertion}\\ \text{for positive series} \end{array}}\left(\left(n_{2}-n_{1}\right)a_{n_{1}}+M\left(n_{2}-n_{1}\right)a_{n_{2}}+M\left(n_{2}-n_{1}\right)a_{n_{3}}+M\left(n_{2}-n_{1}\right)a_{n_{4}}+\ldots\right) $$ $$ =\left(n_{2}-n_{1}\right)\left[a_{1}+M\sum_{k=2}^{n}a_{n_{k}}\right]=\left(n_{2}-n_{1}\right)a_{1}+\left(n_{2}-n_{1}\right)M\sum_{k=2}^{n}a_{n_{k}} $$ $$ \leq\underbrace{\left(n_{2}-n_{1}\right)a_{1}}_{\in\mathbb{R}}+\underbrace{\left(n_{2}-n_{1}\right)M}_{\in\mathbb{R}}\underbrace{\sum_{k=2}^{n}a_{n}}_{\text{Converges}} $$

but I'm not sure I'm allowed to to this.

Answer 1

it's really the same proof as in the wikipedia demonstration :

We establish that,

$$\sum_{n \geqslant n_1} a_n \leqslant \sum_{k \geqslant 2} (n_{k+1} - n_k)a_{n_k} $$

and,

$$\dfrac{1}{M} \displaystyle\sum_{k \geqslant 1} (n_{k+1} - n_k) a_{n_k} +(n_2 - n_1)a_{n_2} \leqslant \sum_{n \geqslant n_1} a_n $$

To see the comparaison, just chose $n_k = 2^k$, then $M= 2$ and the last inequality is like $2\displaystyle\sum_{n \geqslant 1} a_n \leqslant \displaystyle\sum_{k \geqslant 1} 2^n a_{2^n} $

For the first one, we use the upper bound of $a_n$ for $n_{k-1} \leqslant n \leqslant n_k - 1$,

\begin{align}\sum_{n \geqslant n_1} a_n &= \sum_{k \geqslant 2} \sum_{n = n_{k-1}}^{n_k - 1} a_n \\ &\leqslant \sum_{k \geqslant 2} \sum_{n = n_{k-1}}^{n_k - 1} a_{n_{k-1}} \\ &= \sum_{k \geqslant 2} (n_k - n_{k-1}) a_{n_{k-1}} \\ \end{align}

For the second one, we use the lower bound of $a_n$ for $n_{k-1} \leqslant n \leqslant n_k - 1$ and our condition on $(n_k)_k$ (and of course $ M > 0$, otherwise $(n_k)_k$ is constant which is absurd),

\begin{align}\sum_{n \geqslant n_1} a_n &= \sum_{k \geqslant 2} \sum_{n = n_{k-1}}^{n_k - 1} a_n \\ &\geqslant \sum_{k \geqslant 2} \sum_{n = n_{k-1}}^{n_k - 1} a_{n_k} \\ &= \sum_{k \geqslant 2} (n_k - n_{k-1}) a_{n_k} \\ &\geqslant \sum_{k \geqslant 3} \dfrac{1}{M}(n_{k-1} - n_{k-2})a_{n_{k-1}} +(n_2 - n_1)a_{n_2} \\ &= \dfrac{1}{M} \displaystyle\sum_{k \geqslant 1} (n_{k+1} - n_k) a_{n_k} +(n_2 - n_1)a_{n_2} \end{align}

Since we're working with positive $a_n$, we can use series all the time with the convention $\sum a_n = + \infty$ if it diverges. The two inequalities we have provided give the equivalence of convergence !

Edit : To visualize the argument, which is really similar to a series-integral comparaison, you can have a look at the wikipedia page of the Cauchy condensation theorem where they display a graphic which is really helpful to understand this idea of grouping terms and then bounding the same group by above and below to obtain the inequalities in both directions.