Weighted sum with geometric decreasing weights

Question

First time here, but I'm in a sorta challenge. So, let's say we have a sequence $x_i$ with $i=1,2,...,n$ such that $x_i\geq x_j \forall 1 \leq i \leq j \leq n$. Let's define a value $S$ for the sequence: $$S=\frac{1}{5} \sum_{k=1}^n (\frac{4}{5})^{k-1}x_k$$ So, what was making me scratch my head was: what are the general conditions to "edit" a sequence so that its $S$-value will increase? By edit, I mean deleting some element or adding a new one (and if so, what's the optimal range?). What really bugged me was that the $S$ sum can be rewritten as something along the likes of $\frac{\sum_{k=1}^n a_kx_x}{5^n}$ where $\sum_{k=1}^n a_k =5^n-4^n$. If this is true, does that mean that the greater the value of $n$, the harder it is to the $S$-value to increase? Because it gets further from being an normal weighted arithmetic mean? Sorry if I am being too general or sorta not knowing exactly what I am grasping, but any help is welcome. Thanks in advance!

Answer 1

This isn't quite an answer to your question but you can get a lower bound for S.

$S=\frac{1}{5}\sum_{k=1}^{n}{\left(\frac{4}{5}\right)^{k-1}x_{k}} = \frac{1}{4}\sum_{k=1}^{n}{\left(\frac{4}{5}\right)^{k}x_{k}}$

Use Chevyshev's sum inequality to get a minimum value of S:

$S_{min} \geq \frac{1}{4}\left(\frac{1}{n}\sum_{k=1}^{n}{x_{k}}\right)\sum_{k=1}^{n}{\left(\frac{4}{5}\right)^{k}}$

Note that the value in the parentheses is the arithmetic average of all $x_{k}$. Then the lower bound of S becomes:

$S_{min} \geq \frac{\overline{x}}{4}\sum_{k=1}^{n}{\left(\frac{4}{5}\right)^{k}}$

$S_{min} \geq \frac{\overline{x}}{4} \frac{\frac{4}{5}\left(1-\frac{4}{5}^{n}\right)}{1-\frac{4}{5}}$

$S_{min} \geq \overline{x}\left(1-\left(\frac{4}{5}\right)^{n}\right)$

As $n$ increases, $1-\left(\frac{4}{5}\right)^{n}$ increases but $\overline{x}$ decreases. Whether $S_{min}$ increases as $n$ increases depends on the magnitude of change in $\overline{x}$.